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

Question

It is straightforward to see that $f \circ f$ is odd whenever $f$ is odd. Indeed, assuming $f(-x) = -f(x)$ for all $x$, we get

$$ f(f(-x)) = f(-f(x)) = -f(f(x)). $$

Hence, $f \circ f$ is an odd function as well.

My question is a converse of the above statement.

Suppose $f : \mathbb{R} \to \mathbb{R}$ is continuous. If $f \circ f$ is an odd function. What can I say about $f$ itself? Is it odd?

Answer 1

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.