Let $f:[0,1]\to\Bbb{R}$ to be a function satisfying that $$ f(x)=\begin{cases} \frac{f(2x)}{2} &\text{if }x<1/2 \\ \frac{f(2x-1)}{2}+\frac{1}{2} & \text{if } x\ge1/2\end{cases} \qquad \cdots\cdots \qquad (1). $$ then $f(x)=x$ for all $x$?
If $x=\sum_{k=1}^\infty a_k 2^{-k}$ (for each $a_k$ is 0 or 1) then we get $$ f(x)=\frac{1}{2^n}f(2^nx-2^{n-1}a_1-\cdots-2a_{n-1}-a_n)+\sum_{k=1}^n \frac{a_k}{2^k}. $$
If we assume the boundedness of $f$ then we get $f(x)=x$. Besides, we can prove that $f(x)=x$ if $x$ is ratonal number (without assuming the boundedness of $f$). However I don't know there is a unbounded $f$ satisfy (1). In my opinion, the axiom of choice must be used to construct of non-trivial $f$, if it exists. Thanks for any help.
Choose your favorite irrational number $z$ in $(0,1)$ and consider the smallest subset $T(z)\subset(0,1)$ containing $z$ and stable by the functions $u:x\mapsto\frac12x$, $v:x\mapsto\frac12x+\frac12$ and $w:x\mapsto2x-\mathbf 1_{2x\geqslant1}$. Then $T(z)$ can be viewed as an infinite unrooted dyadic tree where the two children of each vertex $x$ in $T(z)$ are $u(x)$ and $v(x)$ and the unique ancestor of $x$ is $w(x)$. One can perturb $f$ on $T(z)$ as follows.
For every vertex $x$ in $T(z)$, there exists a unique pair of nonnegative integers $(i(x),j(x))$ and a unique injective path $(y_0,y_1,\ldots,y_{i(x)+j(x)})$ in $T(z)$ from $y_0=z$ to $y_{i(x)+j(x)}=x$ such that $y_{n}=w(y_{n-1})$ for every $n\leqslant i$ and $y_n$ is $u(y_{n-1})$ or $v(y_{n-1})$ for every $i\lt n\leqslant i(x)+j(x)$. In words, to reach $x$ starting from $z$, one must make $i(x)$ steps towards ancestors, followed by $j(x)$ steps towards some descendants.
Define a function $g$ on $[0,1]$ by $g(x)=c\cdot2^{i(x)-j(x)}$ for every $x$ in $T(z)$, for some $c$, and $g(x)=0$ for every $x$ not in $T(z)$. Then, for every $x$ in $[0,1]$, $g(x)=\frac12g(w(x))$ hence the function $f$ defined by $f(x)=x+g(x)$ solves $$ f(x)=x+\tfrac12g(w(x))=\tfrac12f(w(x))+x-\tfrac12w(x). $$ Since $x-\frac12w(x)=\frac12\mathbf 1_{2x\geqslant1}$, $f$ solves (1). Unless $c=0$, $f$ is unbounded and $f(x)\ne x$ for every $x$ in $T(z)$ (and no choice is involved, so far).
The most general case (with choice) is to start from a partition of $[0,1]\setminus\mathbb Q$ as $\bigcup\limits_{z\in Z}T(z)$, to consider $(i(z,\ ),j(z,\ ))$ and $c(z)$ corresponding to $(i,j)$ and $c$ above, defined on $T(z)$, and to consider the function $f$ defined by $f(x)=x+g(x)$ where $g(x)=c(z)\cdot2^{i(z,x)-j(z,x)}$ for every $x$ in $T(z)$, for some given function $c:Z\to\mathbb R$.
Then $f$ solves (1) and $f(x)=x$ if $x$ is in $[0,1]\cap\mathbb Q$ or in some $T(z)$ with $z$ in $Z$ such that $c(z)=0$. Thus, one can shrink $\{x\in[0,1]\mid f(x)=x\}$ to $[0,1]\cap\mathbb Q$.