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I'm interested in describing solutions to the equation $ f = f \circ \gamma $, where $ f : \mathbb{C} \rightarrow \mathbb{C} $ is a rational function of a given degree, and $ \gamma : \mathbb{C} \rightarrow \mathbb{C} $ is a Mobius transformation. For example, if $ f(z) = z/(z^2+1) $, then $ f(z) = f(1/z) $.

For a fixed $ f $, the set of $ \gamma $ satisfying this equation is a finite subgroup of the automorphism group of the complex projective line, which have been classified: either cyclic, dihedral, or $ A_4 $, $ S_4 $, or $ A_5 $, and you can build rational functions exhibiting all these symmetries. I'm not sure of a good reference for this, but many papers just cite it as well-known.

I have two questions: First, given a rational function $ f $, do there exist nontrivial $ \gamma $ such that $ f = f \circ \gamma $ How can one "read these off" from $ f $? Second, given a Mobius transformation $ \gamma $ and an integer $ d \geq 2 $, does there exist a rational function of degree $ d $ such that $ f = f \circ \gamma $? If so, what is it?

Both of these seem like classical questions and I'm sure they've been completely answered. Just haven't found a good reference for either. I'm also interested in any geometric or alternative method for thinking about this.

Daniel
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    Well, $\gamma$ has to have finite orbits, as $f=f\circ\gamma$ implies that $f=f\circ\gamma^{(n)}$. So, the orbit of any $z_0$ under $\gamma$ has to be contained in a fiber of $f$. For example, if $\gamma(z)=z+1$ there are no rational $f$ with that property. – wisefool Mar 19 '22 at 18:33

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Regarding your comment

"and you can build rational functions exhibiting all these symmetries. I'm not sure of a good reference for this, but many papers just cite it as well-known."

if $R(z)$ is any rational function and if $G$ is a finite group of linear-fractional maps then $ \sum_{g\in G} R(g(z))$ is $G-$ invariant.

(You can also replace the sum by a product to create another invariant rational function.)

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