All solutions for $f(g(x))=g(f(x))$ on $f$

Question

This problem is a general case for Is the identity function the only solution for $f(a^x)=a^{f(x)}$?

This functional equation $f(g(x))=g(f(x))$, $f,g$ invertible, get $f$ over where $x$ is, for example $f(e^x)=e^{f(x)}$, $f(\sqrt{x})=\sqrt{f(x)}$,...., I find out some solutions as:

$f(x)=g(x)$
$f(x)=g^{-1}(x)$
$f(x)=x$
$f(x)=g(g(g(...g(x))))$

I'd like to know if there is any another solution for that and list all solutions.

All of your solutions actually express $f(x) = g^k(x)$ for $k \in \mathbb{Z}$. — gt6989b, Jun 07 '22 at 18:52
If $f(x)=e^{x},$ you don't have $\sqrt{e^x}=e^{\sqrt x}.$ What were you trying to say with the $\sqrt{\cdot}$ statement? — Thomas Andrews, Jun 07 '22 at 19:03
We would need to know something about $g$. In the extremes, $g(x) = 0$ only requires $f(0) = 0$, and $g(x) = x$ has no restrictions at all on $f$. — Brian Moehring, Jun 07 '22 at 19:04
The simple case of $g(x)=x+1$ gives us functions $g$ such that $g(x)-x$ is periodic with period $1.$ There are lot of such functions. — Thomas Andrews, Jun 07 '22 at 19:14
Wait, which is the input function to the question? Is $f$ the given and $g$ unknown? What do you mean by "get $f$ into where $x$ is,...?" — Thomas Andrews, Jun 07 '22 at 19:15
$f(x)=ax$ and $g(x)=bx$, as well as $f(x)=x^n$ and $g(x)=x^m$ also work. — David P, Jun 07 '22 at 19:36
See In which cases are $(f\circ g)(x) = (g\circ f)(x)$?. For polynomials in particular, see the accepted answer under When functions commute under composition. — dxiv, Jun 07 '22 at 20:33
Not common to have this happen with entire functions. Look up the obituary of Baker, https://www.cambridge.org/core/journals/bulletin-of-the-london-mathematical-society/article/irvine-noel-baker-19322001/A152B474E66B853EAB26923FCB1BDB85 Also discussed in KCG, or Iterative Functional Equations by Kuczma, Choczewski, Ger. Right, I forgot the name: permutable functions. — Will Jagy, Jun 07 '22 at 21:02

All solutions for $f(g(x))=g(f(x))$ on $f$

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