Let $C,D$ be two non-compact complex algebraic smooth curves. Suppose that two unramified regular finite maps $p\_1, p\_2: C \rightarrow D$ are given and have the same degree. Is there always an automorphism $\varphi:C \rightarrow C$ such that $p\_1=p\_2\circ \varphi$?



Existence of covering isomorphism
I suppose that "non-compact complex algebraic curve" means complex affine curve.


The following counterexample was proposed by my friend Fedor Pakovich.


Let $D=\mathbf{C}\backslash\{-1,1\}$.
Consider the $4$-th Chebyshev polynomial
$$p\_1(z)=2(2z^2-1)^2-1=8z^4-8z^2+1.$$
It has critical points at $0,\pm1/\sqrt{2}$, with critical values $\pm1$, therefore it defines an
unramified covering
$$p\_1:C\to D,\quad\mbox{where}\quad C=\mathbf{C}\backslash p\_1^{-1}(\{\pm1\}).$$
Now $p\_2(z)=-p\_1(z)$ is another unramified covering
$C\to D$ of the same degree, but evidently $p\_1\neq p\_2\circ\phi.$ (The only non-trivial automorphism of $C$ is $\phi(z)=-z$).


Remarks. 1. This example can be much generalized, of course; one can take any $D$ possessing a non-trivial automorphism $\psi$, then in most cases $p\_1$ (mapping whatever Riemann surfce to $D$)
and $p\_2=\psi\circ p\_1$ will not be related as stated
in the problem.


2. The problem will become harder if under the same assumptions we relax the conclusion to $p\_2=\psi\circ p\_2\circ\phi.$ Fedor and I believe that counterexamples may still exist, but they will be rare.


