Suppose that $s,t \in (\frac{1}{2},1)$ with $t \ne s$. Does there exists a continuous bijection $f \colon [0,1] \to [0,1]$ which simultaneously satisfies the functional equations $$ f(sx) = tf(x) $$ and $$ f(sx + (1-s)) = tf(x) + (1-t) $$
I strongly beleive such a function cannot exist.
Such a function $f$ must satisfy $f(0) = 0$ and $f(1) = 1$. Moreover, it must also satisfy $f(s^n) = t^n$ and $f(1-s^n) = 1-t^n$ for all $n \in \mathbb{N}$. Using this, I can show that for some specific values of $t$ and $s$ such an $f$ cannot exist. For example, if $t=0.9$ and $s = 0.8$ then plotting $(s^n,t^n)$ and $(1-s^n,1-t^n)$ yields the following plot:
Any $f$ satsifying the functional equations would have a graph which goes through each point of the plot. Since $f$ must be stricly increasing we have a contradiction. On the other hand, if $t=0.6$ and $s = 0.55$ then the corresponding plot is given by:
Here there is no clear obstruction to injectivity.
Edit: At the suggestion of mathworker21, plotting more points in the "graph" of $f$ for $t=0.6$ and $s = 0.55$ yeilds the following plot:
Again such and $f$ cannot be injective, but it is not clear how to turn this into a proof.
Note: I previously made a post Solutions to the functional equations $f(sx) = tx$ and $f(sx + (1-s)) = tf(x) + (1-t)$ on $[0,1]$ where the question I asked in the body contained a typo. Since I already awarded the correct answer for the incorrect question, I thought I should make a new post to not take away from the answers that were contributed.
Let $f$ be a continuous bijection from $\left[0, 1\right]$ to $\left[0, 1\right]$ such that for $x \in \left[0, 1\right]$ holds \begin{align} f\!\left(sx\right) &= tf\!\left(x\right) \label{constr 1} \tag{1} \\ f\!\left(sx + 1 - s\right) &= tf\!\left(x\right) + 1 - t \label{constr 2} \tag{2} \end{align} for some fixed pair $\left(s, t\right) \in \left(2^{-1}, 1\right)^{2}$, $s \neq t$. If such function exists, then it's inverse $f^{-1}:\left[0, 1\right] \rightarrow \left[0, 1\right]$ is continuous bijection, which satisfies \begin{align} f^{-1}\!\left(tx\right) &= sf^{-1}\!\left(x\right) \\ f^{-1}\!\left(tx + 1 - t\right) &= sf^{-1}\!\left(x\right) + 1 - s \end{align} That's why without loss of generality we asume that $s > t$. I will work with modified equation \eqref{constr 2}, where substitution $x \rightarrow 1 - x$ is used \begin{equation} f\!\left(1 - sx\right) = tf\!\left(1 - x\right) + 1 - t \label{constr 2'} \tag{2'} \end{equation} Set of finite strictly increasing sequence $\left(n_k\right)$ of numbers $n_{k} \in \mathbb{N} \cup \left\{0\right\}$, $0 \leqslant k \leqslant l$ we denote as $\mathfrak{S}_{l}$. For $\left(n_{k}\right) = \mathfrak{s} \in \mathfrak{S}_{l}$ and $x \in \left(0, 1\right)$ we define next series \begin{equation} \mathcal{E}\!\left(x;\mathfrak{s}\right) = \sum\limits_{k \,=\, 0}^{l} (-1)^{k}x^{n_{k}} \end{equation} When sequence $\left(n_k\right)$ is infinite and fixed, but also strictly increasing, we will denote as $\mathcal{S}_{l}\!\left(x\right)$ for partial sums of series \begin{equation} \mathcal{S}\!\left(x\right) = \sum\limits_{k \,=\, 0}^{\infty} (-1)^{k}x^{n_{k}} \end{equation} Note that $\mathcal{S}_{l}\!\left(x\right)$ and $\mathcal{E}\!\left(x;\mathfrak{s}\right)$ lies in $\left[0, 1\right]$ and $\mathcal{S}\!\left(x\right)$ always finite as series absolutely converges \begin{equation} \sum\limits_{k \,=\, 0}^{\infty} x^{n_{k}} \leqslant \sum\limits_{k \,=\, 0}^{\infty} x^{k} = \dfrac{1}{1 - x} \end{equation}
Let $y \in \left(0, 1\right) \backslash \, \mathbb{E}\left(\mathcal{E}\!\left(x;\cdot\right)\right)$. We construct sequence $\left(n_k\right)$ inductively with next properties: \begin{align} & (-1)^{l}\mathcal{S}_{l}\!\left(x\right) &&> y \label{S one} \tag{3} \\ & (-1)^{l}\mathcal{S}_{l}\!\left(x\right) - x^{n_{l}} + x^{n_{l} + 1} &&< y \label{S two} \tag{4} \end{align} Sequence of points $\left(x^{n}\right)$, $n \geqslant 0$ represents a "point" partition of an $\left(0, 1\right)$. Under "point" partition $\mathbf{P}$ of $\left(0, 1\right)$ I mean sequence of points, such that set of intervals $\left(a_i, a_{i + 1}\right)$ $\left(\text{or } \left(a_{i + 1}, a_{i}\right)\right)$ creates a partition of $\left(0, 1\right) \backslash\, \mathbf{P}$.
So there exists unique $n_{0} \in \mathbb{N} \cup \left\{0\right\}$ such \begin{align} \mathcal{S}_{0}\!\left(x\right) - x^{n_{0}} + x^{n_{0} + 1} &< y < \mathcal{S}_{0}\!\left(x\right) \\ x^{n_{0}+1} &< y < x^{n_{0}} \end{align} and base of induction is proved. To prove induction step $(l \rightarrow l + 1)$ we use that $x \geqslant 1 - x$: \begin{align} (-1)^{l}\mathcal{S}_{l}\!\left(x\right) - x^{n_{l} + 1} &= (-1)^{l}\sum\limits_{k \,=\, 0}^{l - 1} (-1)^{k}x^{n_{k}} + x^{n_{l}}\left(1 - x\right) \\ &\leqslant (-1)^{l}\sum\limits_{k \,=\, 0}^{l - 1} (-1)^{k}x^{n_{k}} + x^{n_{l} + 1} \\ &= (-1)^{l}\mathcal{S}_{l}\!\left(x\right) - x^{n_{l}} + x^{n_{l} + 1} \\ &< y \end{align}
Last inequalities shows that set of all $n > n_{l}$ such that \begin{equation} (-1)^{l}\mathcal{S}_{l}\!\left(x\right) - x^{n_{l} + 1} < y \end{equation} is not empty. Combining this with \eqref{S one} and with monotonicity of $x^{n}$ we conclude that there unique $n_{l + 1}$ such \eqref{S one}, \eqref{S two} holds true for $l+1$. Taking limit in \eqref{S one} of an even/odd subsequences, we get that $\mathcal{S}\!\left(x\right) = y$.
Partial sums $\mathcal{S}_{l}\!\left(x\right)$ are elements of $\mathbb{E}\left(\mathcal{E}\!\left(x;\cdot\right)\right)$, which means that $\mathcal{E}\!\left(x;\cdot\right)$ dense in $\left(0, 1\right)$.
We can restate one part of Lemma 1 as follows
From \eqref{constr 1} and \eqref{constr 2'} it's provable via induction that for natural $n$ and $x \in \left[0, 1\right]$ \begin{align} f\!\left(s^{n}x\right) &= t^{n}f\!\left(x\right) \label{pow 1} \tag{6} \\ f\!\left(1 - s^{n}x\right) &= t^{n}f\!\left(1 - x\right) + 1 - t^{n} \label{pow 2} \tag{7} \end{align}
We will prove this claim by induction on $l$. Base of induction follows from next sequence of transformations \begin{align} f\!\left(\mathcal{E}\!\left(s;\mathfrak{s}\right) - s^{n_{0} + p}x\right) &= \mathcal{E}\!\left(t;\mathfrak{s}\right) - t^{n_{0} + p}\Big(1 - f\!\left(1 - x\right)\!\Big) \\ f\!\left(s^{n_{0}} - s^{n_{0} + p}x\right) &= t^{n_{0}} - t^{n_{0} + p}\Big(1 - f\!\left(1 - x\right)\!\Big) \\ f\!\left(1 - s^{p}x\right) &= 1 - t^{p}\Big(1 - f\!\left(1 - x\right)\!\Big) \end{align} Let us prove case $2l \rightarrow 2l + 1$. For $\left(n_{k}\right) = \mathfrak{s}' \in \mathfrak{S}_{2l + 1}$ we organize $n_{0}, \dots, n_{2l}$ in new sequence $\mathfrak{s}$. Thus \begin{equation} \mathcal{E}\!\left(x;\mathfrak{s}'\right) = \mathcal{E}\!\left(x;\mathfrak{s}\right) - x^{n_{2l + 1}} \end{equation}
Take $p = n_{2l + 1} - n_{2l}$ and $x \rightarrow 1 - s^{p'}x$, $p' \geqslant 0$ \begin{align} f\Big(\mathcal{E}\!\left(s;\mathfrak{s}\right) - s^{n_{2l + 1}}\left(1 - s^{p'}x\right)\!\Big) &= f\Big(\mathcal{E}\!\left(s;\mathfrak{s}\right) - s^{n_{2l + 1}} + s^{n_{2l + 1} + p'}x\!\Big) \\ &= f\!\left(\mathcal{E}\!\left(s;\mathfrak{s}'\right) + s^{n_{2l + 1} + p'}x\right) \\ &= \mathcal{E}\!\left(t;\mathfrak{s}\right) - t^{n_{2l + 1}}\Big(1 - f\!\left(s^{p'}x\right)\!\Big) \\ &= \mathcal{E}\!\left(t;\mathfrak{s}'\right) + t^{n_{2l + 1} + p'}f\!\left(x\right)\! \end{align} For case $2l + 1 \rightarrow 2l + 2$ we do the same. With properly selected parameters we have \begin{align} f\Big(\mathcal{E}\!\left(s;\mathfrak{s}\right) + s^{n_{2l + 2}}\left(1 - s^{p'}x\right)\!\Big) &= f\Big(\mathcal{E}\!\left(s;\mathfrak{s}\right) + s^{n_{2l + 2}} - s^{n_{2l + 2} + p'}x\!\Big) \\ &= f\!\left(\mathcal{E}\!\left(s;\mathfrak{s}'\right) - s^{n_{2l + 2} + p'}x\right) \\ &= \mathcal{E}\!\left(t;\mathfrak{s}\right) + t^{n_{2l + 2}}f\!\left(1 - s^{p'}x\right)\! \\ &= \mathcal{E}\!\left(t;\mathfrak{s}'\right) - t^{n_{2l + 1} + p'}\Big(1 - f\!\left(1 - x\right)\!\Big) \end{align} As consequence we get that for any $l \geqslant 0$ and $\mathfrak{s} \in \mathfrak{S}_{l}$ \begin{equation} f\!\left(\mathcal{E}\!\left(s;\mathfrak{s}\right)\right) = \mathcal{E}\!\left(t;\mathfrak{s}\right) \label{f fin} \tag{8} \end{equation} In addition, with Lemma 1 and continuity of $f$ we conclude that for any strictly increasing sequence $\left(n_k\right)$ holds true \begin{equation} f\!\left(\sum\limits_{k \,=\, 0}^{\infty} (-1)^{k}s^{n_{k}}\right) = \sum\limits_{k \,=\, 0}^{\infty} (-1)^{k}t^{n_{k}} \label{f inf} \tag{9} \end{equation} As we can see from \eqref{f fin}, \eqref{f inf}, it doesn’t matter to us that the sequence $\left(n_{k}\right)$ is finite or infinite. Therefore, we will not pay attention to this and will simply write the index $k$ at the bottom of the sums of type like in \eqref{repr}.
Claim enough to prove for $x \in \left[0, 2^{-1}\right]$. From Lemma 1 we know, that it possible for $x \in \left[0, 2^{-1}\right]$ to construct $\left(n_{k}\right)$ such that $n_{0} \geqslant 1$ $\left(\text{as } s^{1} > 2^{-1}\right)$ and \begin{equation} x = \sum\limits_{k} (-1)^{k}s^{n_{k}} \end{equation} Then \begin{align} f\!\left(1 - x\right) + f\!\left(x\right) &= f\!\left(1 - \sum\limits_{k} (-1)^{k}s^{n_{k}}\right) + f\!\left(\sum\limits_{k} (-1)^{k}s^{n_{k}}\right) \\ &= 1 - \sum\limits_{k} (-1)^{k}t^{n_{k}} + \sum\limits_{k} (-1)^{k}t^{n_{k}} \\ &= 1 \end{align}
As consequence we get that $f\!\left(2^{-1}\right) = 2^{-1}$.
Assume that \eqref{low x} is not true. It means that for any $\varepsilon > 0$ there exists $0 < x_{\varepsilon} < \varepsilon$ such that $f\!\left(x\right) \geqslant x$. But we have $f\!\left(s^{n}\right) = t^{n} < s^{n}$ for natural $n$ and $s^{n} \rightarrow 0$. Thus we can find a monotone sequence $x_{\varepsilon_{1}}, s^{n_{1}}, x_{\varepsilon_{2}}, s^{n_{2}}, \dots$ which converges to zero and for natural $k$ satisfies \begin{align} f\!\left(x_{\varepsilon_{k}}\right) &\geqslant x_{\varepsilon_{k}} \\ f\!\left(s^{n_{k}}\right) &< s^{n_{k}} \end{align}
Applying Weierstrass theorem for every interval $\left[ s^{n_{k}}, x_{\varepsilon_{k}}\right]$ we find a monotone, converging to zero sequence $\left(y_{n}\right)$ of fixed points of $f$. Hence $f$ is strictly monotonous, difference $y_{n} - s^{n_{0}}$ has the same sign with $y_{n} - t^{n_{0}}$. To see this, for example, when $y_{n} < s^{n_{0}}$, just go to image of $f$. From here we can conclude that for $n \geqslant p$ \begin{equation} \left|\log_{s} y_{n} - \log_{t} y_{n}\right| \leqslant 1 \label{s, t ineq} \tag{11} \end{equation} where $p$ is selected so that $y_{p} < s$. Indeed, for $m = \left\lfloor \log_{s} y_{n} \right\rfloor \in \mathbb{N} $ \begin{equation} t^{m + 1} < s^{m + 1} < y_{n} < t^{m} < s^{m} \end{equation} which means that \begin{equation} m < \log_{s} y_{n}, \log_{t} y_{n} < m + 1 \end{equation} and \eqref{s, t ineq} is true. Note that all inequalities a strict, because $s > k$ and $y_{n} = s^{k}$ for some $k$ implies $y_{n} = t^{k}$. Now rewrite \eqref{s, t ineq} in the next form \begin{align} \left|1 - \dfrac{\log_{t} y_{n}}{\log_{s} y_{n}}\right| &\leqslant \dfrac{1}{\left|\log_{s} y_{n}\right|} \\ \left|1 - \log_{t} s\right| &\leqslant \dfrac{1}{\left|\log_{s} y_{n}\right|} \end{align} Taking limit as $n \rightarrow \infty$, we come to the conclusion that $\log_{t} s = 1$ or $t = s$, which contradicts with initial condition.
Set of $\beta > 0$ such that on $\left(0, \beta\right)$ holds \eqref{low x} is nonempty and bounded: $b \leqslant 2^{-1}$ as $f\!\left(2^{-1}\right) = 2^{-1}$, so it's has a supremum $x_{0}$. From local properties of continuous functions, we have that \begin{align} f\!\left(x\right) &< x,\quad x \in \left(0, x_{0}\right) \\ f\!\left(x_{0}\right) &= x_{0} \end{align} and $x_{0} \leqslant 2^{-1} < s$. Now we find $n \in \mathbb{N}$ such \begin{equation} s^{n + 1} < x_{0} < s^{n} \end{equation} One more time, there are strict inequalities as $x_{0}$ is a fixed point, and points $s^{k}$, $k \geqslant 1$ are not. From Claim 1 we have that for $x \in \left[0, s^{n+1}\right]$ \begin{align} f\!\left(s^{n} - s^{n+1} + x\right) &= t^{n} - t^{n+1} + f\!\left(x\right) \\ &= f\!\left(s^{n} - s^{n+1}\right) + f\!\left(x\right) \\ &< f\!\left(s^{n} - s^{n+1}\right) + x \end{align} Now take $x = x_{0} + s^{n+1} - s^{n} \in \left[0, s^{n+1}\right]$ to have \begin{align} f\!\left(x_{0}\right) &= x_{0} \\ &< f\!\left(s^{n} - s^{n+1}\right) + x_{0} + s^{n+1} - s^{n} \\ s^{n} - s^{n+1} &< f\!\left(s^{n} - s^{n+1}\right) \end{align} but $s^{n} - s^{n+1} = s^{n}\left(1 - s\right) < s^{n + 1}$, and we must have \begin{equation} f\!\left(s^{n} - s^{n+1}\right) < s^{n} - s^{n+1} \end{equation} by Claim 3. This contradiction completes the proof of nonexistence of such $f$.