Decide whether or not there exists $f(f(x))=5x-\lfloor 5x\rfloor$ 
Decide whether or not there exists  $f;[0,1)\to [0,1)$ such 
  $$f(f(x))=5x-\lfloor 5x\rfloor=\{5x\}$$

Today I suddenly thought of an interesting question, but I have not found any function of symbolic conditions so far.

Yes, it it exists as the following example shows.The function $\{5x\}$ in the interval $[0,1)$ is explicitly defined by
$$\begin{equation}\{5x\}=\begin{cases}5x\space \text{ for }0\le x\lt 0.2\\5x-1\space \text{ for }0.2\le x\lt 0.4\\5x-2\space \text{ for }0.4\le x\lt 0.6\\{5x-3}\space \text{ for }0.6\le x\lt 0.8\\5x-4\space \text{ for }0.8\le x\lt 1\end{cases}\end{equation}$$ so we can calculate an example of $f(x)=\sqrt5\space x+b$ such that $f(f(x))$ coincides with the funcion above. Calculation gives
$$\begin{equation}f(x)=\begin{cases} \sqrt 5\space x\quad\text{ for }0\le x\lt 0.2\\\\\sqrt 5\space x-\dfrac{\sqrt5-1}{4} \text{ for }0.2\le x\lt 0.4\\\\\sqrt 5\space x-\dfrac{(\sqrt5-1)}{2}\space \text{ for }0.4\le x\lt 0.6\\\\\sqrt 5\space x-\dfrac{3(\sqrt5-1)}{4}\space \text{ for }0.6\le x\lt 0.8\\\\\sqrt 5\space x-(\sqrt5-1)\space \text{ for }0.8\le x\lt 1\end{cases}\end{equation}$$ It follows
 $$\color{red}{f(f(x))=\{5x\}}$$ as it is easy to verify.
