Existence of covering isomorphism

Question

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 $p_1=p_2\circ \varphi$?

Answer 1

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.