If $f \circ f$ is odd, then is so $f$?

calculus even-and-odd-functions

Define $f(x)=\begin{cases}0&x\leq0\\-x&x>0\end{cases}$.

Then $f(f(-x))=-f(f(x))=0$ but $f(-x)\neq-f(x)$, except at $x=0$. Hence we have an odd $f(f(x))$ which doesn't imply an odd $f(x)$.

Note that $f(f(x))$ is in fact both even and odd. This answer was inspired in part by user @Henry_Lee.

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