Find all continuous $f:[0,1] \rightarrow [0,1]$ such that $f(1-f(x))=f(x)$.
Let $m,M$ be the minimum and maximum $f$ achieves on $[0,1]$ (there are such since f is continuous). From the intermediate value theorem, for each $m\leq y \leq M$ there is an $x\in [0,1]$ such that $f(x)=y$, so $f(1-y)=f(1-f(x))=f(x)=y$. This shows that if $m \leq y\leq M$ then $f(1-y)=y$.
for the $ [0,1] - [1-M, 1-m] $ you can extend $f$ any way you want as long as its range is in $[m,M]$.