The question: Let $\sigma$ be a permutation of $\{1,\dots,n-1\}$. Show that there exists a continuous function $f:[0,1]\to[0,1]$ with a $n$-periodic point $x$ such that $x<f^{\sigma(1)}(x)<\dots<f^{\sigma(n-1)}(x)$.
The proof should be very basic, as the question was posed after a brief explanation of Sharkovskiis theorem. My thoughts: take any continuous function with a $n$-periodic point $u$. Then you can simply take $x=\min_{k\in\{1,\dots,n-1\}}f^{\sigma(k)(u)}$. The obvious problem is that this method gives $\sigma$ belonging to $f$ and $x$ instead of the other way around.
I was wondering whether it is possible to find a conjugation $\pi$ between $C[0,1]$ and $\Sigma^{n-1}$, the space containing all permutations of $\{1,\dots,n-1\}$.
You're overthinking this problem. You can just write down such a function. Take any $n-1$ points $a_1, \ldots, a_{n-1} \in (0,1)$ with indices chosen so that $$a_1 < a_{\sigma(1)} < a_{\sigma^2(1)} < \ldots < a_{\sigma^{n-1}(1)}.$$ Define $f$ so that $f(0)=0$, $f(1)=1$ and $f(a_i) = a_{\sigma(i)}$ and everything in between these values is just piece-wise linearly interpolated between these points.
Such a function is obviously continuous and $a_1$ is the periodic point that is needed. It's easy to check that it satisfies $a_1 < f^{\sigma(1)}(a_1) < \cdots < f^{\sigma(n-1)}(a_1)$ because $f^i(a_1) = a_{\sigma^i(1)}$.