Is it possible to find out two functions $f, g:\mathbb{R}\to\mathbb{R}$ such that $f(g(x))=x$ and $g(f(x))=-x$?

Question

This is related to Prove that there exists $f,g : \mathbb{R}$ to $\mathbb{R}$ such that $f(g(x))$ is strictly increasing and $g(f(x))$ is strictly decreasing.

But according to the proof of Ewan Delanoy, you must use a $p(x)=kx$, $q(x)=-kx$, where $k\neq1$, otherwise the iterative group is trivial, then the proof failed. I think it is still possible to construct such $f,g$ that their compositions are exactly $y=\pm x$.

Answer 1

Consider $$ g(x)=g(f(g(x)))=-g(x) $$

Answer 2

If $f(g(x))=x$, then $f$ is surjective and $g$ is injective. And if $g(f(x))=-x$, then $g$ is surjective and $f$ is injective. Therefore, $f$ and $g$ are bijections and then it follows from $f(g(x))=x$ that $g=f^{-1}$. But this contradicts the fact that $g(f(x))=-x$.