Find all functions $f:[0,1] \rightarrow[0,1]$ such that for all $0 \leqslant x \leqslant 1, f(2 x-f(x))=x$.


Keys: function change, iterates of a function.

Let $g(x):=2 x-f(x)$ for all $x$. We see that $f \circ g$ is the identity, which is a bijective function, implying that $f$ and $g$ are bijective (this is a useful classical fact). We note that $g$ takes values in $[0,1]$ since $f(g(x))$ is well-defined for all $x$. We denote, for $n \geqslant 1, g^{n}=g \circ \ldots \circ g$ (composing $g$ $n$ times). It is easily seen by induction that $g^{n}(x)=n g(x)-(n-1) x=n(g(x)-x)+x$. If there exists $x_{0}$ such that $g\left(x_{0}\right)-x_{0} \neq 0$, then for $n$ large enough $\left|g^{n}\left(x_{0}\right)\right|>1$, but $g$ takes values in $[0,1]$ so does $g^{n}$, contradiction. Therefore, $g(x)=x$ for all $x$, hence $f(x)=x$ for all $x$. Conversely, this function is a solution.