a continuous function satisfying $f(f(f(x)))=-x$ other than $f(x)=-x$ Is there a non-trivial solution to the functional equation $f(f(f(x)))=-x$ where $f$ is a continuous function defined on $\mathbb{R}$ ? Also, what about the general one $f^n(x)=-x$ where $f^n$ is $f$ composed with itself $n$ times and $n$ is odd. 
In general, is there a theory about continuous solutions to $f^n(x)=g(x)$ where $g$ is a fixed continuous function. The only thing that I found was about the solutions of $f^2(x)=g(x)$, i.e, "square roots" in the sense of composition.
Thanks.
Edit : I forgot to mention that $n$ is supposed to be odd, sorry for the inconvenience.

As Henry points out in the comments, when $n$ is odd $f(x)=-x$ obviously works. 
When $n$ is even, there is no such function. To see this, note the following properties must hold:


*

*$f$ is a bijection, and hence stricly monotonic.

*$f(0)=0$. Indeed, if $f(0)=a$, then $f^n(a) = f^{n+1}(0) = f(f^n(0)) = f(0) = a$, so $a=0$ since $0$ is the only fixed point of $-x$.

*From the previous two properties, we must have that $f$ is either increasing, in which case $f(x)$ has the same sign as $x$ for every $x$, or decreasing, in which case $f(x)$ has the opposite sign as $x$ for every $x$. Either way, $f^n(x)$ and $x$ share the same sign, and so $f^n(x) \neq -x$.

