Solution:
The answer is the 32 values $0, \frac{1}{31}, \frac{2}{31}, \ldots, \frac{30}{31}, 1$.

If $x<0$, then $f(x)=2 x<x$ so the sequence $x, f(x), f(f(x)), \ldots$ is strictly decreasing and cannot return to $x$.

If $x>1$, similarly $f(x)=2 x-1>x$ so the sequence $x, f(x), f(f(x)), \ldots$ is strictly increasing and cannot return to $x$.

If $x=1$, then $f(x)=1$ and we have $1$ as a solution.

Finally, we assume that $0 \leq x<1$, so $0 \leq f(x)<1$ as well. For simplicity let $x_{0}=x$ and $x_{n+1}=f\left(x_{n}\right)$ so the equation we are trying to solve is $x_{5}=x_{0}$. Note that $2 x_{n}-x_{n+1}$ is an integer (either $0$ or $1$) for each $n$, so
$$32 x_{0}-x_{5}=16\left(2 x_{0}-x_{1}\right)+8\left(2 x_{1}-x_{2}\right)+4\left(2 x_{2}-x_{3}\right)+2\left(2 x_{3}-x_{4}\right)+\left(2 x_{4}-x_{5}\right)$$
must be an integer as well. If we assume $x_{5}=x_{0}$ we deduce that $31 x_{0}$ is an integer. Conversely, if $31 x_{0}$ is an integer, then $x_{5}-x_{0}$ is an integer and this integer must be $0$ because $0 \leq x_{0}, x_{5}<1$. Thus the solutions in this range are exactly the multiples of $\frac{1}{31}$: $0, \frac{1}{31}, \frac{2}{31}, \ldots, \frac{30}{31}$.