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Assume that $f(z)$ is a holomorphic function that sends some open and connected set $G$ to itself. Assume $f$ has a single fixed point $z_0$. Assume $f(f(...(n\,times)...f(z))) = f^{\circ n}(z) \to z_0$ as $n \to \infty$ for all $G$. By consequence $|f'(z_0)| < 1$. Let us additionally assume $f'(z_0) \neq 0$.

This allows us to conclude there is a Schröder function $\Psi : G \to \Psi(G)$ such that

$$\Psi(f(z)) = f'(z_0)\Psi(z)$$ $$\Psi'(z_0) = 1$$

which also means $\Psi$ is injective in a small neighborhood about $z_0$. Let us take $g$ which satisfies the exact same criteria as $f$. It has a Schröder function too, say $\Phi:G \to \Phi(G)$.

Is there a way to know whether there exists a map $\chi: \Psi(G) \to \Phi(G)$ such that $\chi(\Psi(z)) = \Phi(z)$?

It follows that such a function exists if we shrink $G$, to say $U$ a tiny neighborhood of $z_0$. We can easily take $\chi(z) = \Phi(\Psi^{-1}(z))$ where $\chi: \Psi(U) \to \Phi(U)$. But in general, what is the maximal domain this works? When does it begin to fail as we let $U$ get bigger? Why?

Does it happen for the obvious reason that $\Psi^{-1}$ is no longer defined as we grow $U$? This is to mean $\Psi$ must be biholomorphic from $G \to \Psi(G)$ for $\chi$ to exist. But, I ruminate that it must be more complicated than that. Clearly if $g = f(f(z))$ then there does exist such a map. Simply take the identity function $\chi = \text{Id}$. It follows similarly if $g$ is in some other orbit of $f$. This follows regardless if $\Psi$ is biholomorphic.

I wonder if maybe these are the only particular cases. That it only really happens if $\Psi$ is invertible or $\chi = \text{Id}$. To make sense of this statement it might help to restrict ourselves to rational maps taking $\hat{\mathbb{C}}$ to itself, and letting $G$ be the immediate basin of attraction about a geometric fixed point $z_0$. If there is some $g$ on $G$, and it has its own Schroder function $\Phi: G \to \Phi(G)$, assuming there exists $\chi$ between $f$ and $g$: then $\Psi$ is biholomorphic, or $\chi = \text{Id}$.

But then, I think mathematics is too ingenious for that, there's got to be more to the story.

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    $\begingroup$ There are various things that are odd in your question. Firstly, the Schröder function will not map $G$ to itself. Secondly, what kind of "map" are you speaking of? Should it be a homeomorphism? Thirdly, are you assuming anything about the mapping of the function of the invariant domain to itself? Should this be global (e.g., a proper map of the domain onto itself) in some sense, so that the linearisers naturally map to the full complex plane? $\endgroup$ – Lasse Rempe-Gillen Mar 1 '17 at 18:53
  • $\begingroup$ If you are speaking about global linearisers, let's say for a rational function, related by an affine map, then you are effectively asking: for which different functions can a given Schröder function arise? I believe this should be answerable from known results. Indeed, for a given multiplier, the map is uniquely determined by the Schröder function (up to conformal conjugacy). And the possible multipliers are constrained by the positions of the critical values of $\Phi$. So maybe this happens only for iterates of a common base function, but I have not given it sufficient thought. $\endgroup$ – Lasse Rempe-Gillen Mar 1 '17 at 19:01
  • $\begingroup$ @LasseRempe-Gillen I know $\Psi$ doesn't map $G \to G$, it maps $G \to \Psi(G)$ (whatever that is, a domain about zero; it could be all of $\mathbb{C}$). The function $\chi$ doesn't necessarily have to be a homeomorphism, just a holomorphic function that satisfies $\chi(\Psi(z)) = \Phi(z)$ for $\Phi$ a different Schroder function. The main interest is solving this functional equation; and in what instances can it be solved. As to your last line, that's what I was thinking for rational maps $f: \hat{\mathbb{C}}\to\hat{\mathbb{C}}$ about the basin of some fixed point $z_0$. $\endgroup$ – user78249 Mar 1 '17 at 19:27
  • $\begingroup$ Thank you for the clarification. It would be good to edit the question to reflect this. You may also wish to reconsider your title, as "conjugating" usually implies a) a dynamical equivalence (whereas here you are only postcomposing), and b) a bijection. $\endgroup$ – Lasse Rempe-Gillen Mar 2 '17 at 5:29
  • $\begingroup$ @LasseRempe-Gillen You're right. I shouldn't have used "conjugated" so loosely. I was thinking less mathematically, and too linguistically. I'll try to edit the question to be clearer. Thank you for your interest and comments. $\endgroup$ – user78249 Mar 2 '17 at 6:25

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