Solutions for a functional equation

Question

What are the functions (complex or real) that accept these properties: $(f \circ f \circ f \circ f) (x) = x$ but $(f \circ f) (x) ≠ x$? I mean that these functions are not self-inverse but their compositions with themselves are self-inverse. I have not any clue for solving this functional equation.

Answer 1

I presume you want a function $f: \mathbb R \to \mathbb R$ such that $f(f(f(f(x)))) = x$ for all $x$ but $f(f(x)) \ne x$ for at least some $x$. The first equation implies that $f$ is one-to-one and onto. It's not hard to show that there is no continuous solution. On the other hand, it's not hard to find discontinuous solutions. Just partition $\mathbb R$ into four sets $A, B, C, D$, let $f$ be arbitrary one-to-one maps of $A$ onto $B$, $B$ onto $C$, $C$ onto $D$, and on $D$ let $f: x \mapsto y \in A$ where $f(f(f(y))) = x$.