Suppose you've got a Riemann surface $E$, and two topological covers $X,Y\rightarrow E$. Suppose $X,Y$ are nonisomorphic topological covers of $E$, then would you expect $X,Y$ as Riemann surfaces (with their unique complex structure coming from the covering map) to be nonisomorphic?
I know there are some counterexamples. I was just wondering if these counterexamples are rare, and in what sense (and why) they're rare.
There's a remarkable theorem of Margulis that pertains to your question. Let $G$ be a semisimple Lie group (in your case, $PSL_2(\mathbb{R})$), and let $\Gamma$ be an irreducible lattice in $G$. The commensurator of $\Gamma$ is defined as $$Comm(\Gamma)=\{ g\in G\ |\ [\Gamma: g\Gamma g^{-1}\cap \Gamma]<\infty\},$$ the subset of $G$ which "almost" normalizes $\Gamma$. For example, if $\Gamma <\Gamma'$ a larger lattice, then $\Gamma' \leq Comm(\Gamma)$ (notice that if $\Gamma$ is not normal in $\Gamma'$, then $g\Gamma g^{-1}\neq \Gamma$ for some $g\in \Gamma'$).
Margulis' theorem implies that either $Comm(\Gamma)$ is discrete, or $\Gamma$ is arithmetic. If $\Gamma$ is non-uniform (and $G=PSL_2(\mathbb{R})$), so $\mathbb{H}^2/\Gamma$ is finite volume but non-compact, then $\Gamma$ is commensurable with $PSL_2(\mathbb{Z})$, in other words, there is $g\in PSL_2(\mathbb{R})$ so that $[\Gamma: \Gamma \cap g PSL_2(\mathbb{Z}) g^{-1}] <\infty$. If $\Gamma$ is uniform, so $\mathbb{H}^2/\Gamma$ is a compact Riemann surface, then the description is a bit more complicated, but basically $\Gamma$ is commensurable with a discrete group defined by integral automorphisms of a ternary quadratic form of signature $(2,1)$ defined over a totally real number field, and which is definite at all other infinite places, using the isogeny $PGL_2(\mathbb{R})\cong O(2,1;\mathbb{R})$ (I think in the number theory lingo these are known as Shimura curves).
Now, suppose in your question that $X$ and $Y$ are finite area (in the unique complete hyperbolic metric given by the uniformization theorem; I can't say much of anything about the infinite area case), so define finite-sheeted covers of $E$. We may associate a torsion-free lattice $\Gamma < PSL_2(\mathbb{R})$ so that $E=\mathbb{H}^2/\Gamma$, and finite-index subgroups $\chi, \gamma < \Gamma$ so that $\mathbb{H}^2/\chi = X, \mathbb{H}^2/\gamma=Y$. Suppose $X$ and $Y$ are isomorphic Riemann surfaces, then there is $g\in PSL_2(\mathbb{R})$ so that $\chi = g\gamma g^{-1}$. Therefore $\Gamma > \chi = g \gamma g^{-1} < g \Gamma g^{-1}$, so $[\Gamma : \Gamma \cap g\Gamma g^{-1}] <\infty$, so $g\in Comm(\Gamma)$.
Thus, by Margulis' theorem, either $\Gamma$ is arithmetic, or $\Gamma$ is non-arithmetic, and therefore $\Lambda=Comm(\Gamma)$ is discrete. Then $\mathbb{H}^2/\Lambda$ is a finite-area hyperbolic orbifold $\mathcal{O}_\Lambda$, and one has that $g\Lambda g^{-1} =\Lambda$. This implies that $E\to \mathcal{O}_\Lambda$ is a finite-sheeted orbifold cover, and the two covers $X,Y\to E\to \mathcal{O}_\Lambda$ are equivalent (irregular) covers of $\mathcal{O}_\Lambda$. Notice that if $\mathcal{O}_\Lambda$ has non-trivial modulus (so it's not a turnover), then this gives rise to parameter space of such covers of Riemann surfaces of the same genus as $E$.
In the case $\Gamma$ is arithmetic, $Comm(\Gamma)$ is much larger. For example, $Comm(PSL_2(\mathbb{Z}))\cong PGL_2(\mathbb{Q})$. Such pairs of covers then are an important area of study in the theory of autormorphic forms, since they give rise to Hecke operators. However, for a fixed genus, there are only finitely many arithmetic Riemann surfaces of that genus (see e.g. Long-Reid for a more general result for orbifolds), so in particular there are only finitely many such examples for which $X$ and $Y$ have bounded area. There is (in principle) a complete arithmetic prescription for how such covers may occur in this case as well which is determined by the description of the commensurator.
A cover $\alpha:X\rightarrow E$ induces a map between Teichmuller spaces:
$\alpha^*:Teich(E)\rightarrow Teich(X)$.
Let $\beta:X\rightarrow E$ be another cover with induced map
$\beta^*:Teich(E)\rightarrow Teich(X)$.
If $\alpha$ and $\beta$ are holomorphic then the induced maps $\alpha^*$ and $\beta^*$ send the basepoint of $Teich(E)$ to the basepoint of $Teich(X)$.
For $\tau\in Teich(E)$, an isomorphism between the surfaces represented by $\alpha^*(\tau)$ and $\beta^*(\tau)$ is given by an element of the Mapping Class Group. Since Teichmuller space is connected and since the Mapping Class Group acts properly discontinuously, it follows that $\beta^* = \phi^*\circ\alpha^*$ where $\phi:X\rightarrow X$ represents a particular given mapping class. Since $\phi^*$ fixed the basepoint, we may as well assume that $\phi$ is a conformal automorphism of $X$, and compose it with $\beta$ to yield a cover $\hat{\alpha}:X\rightarrow E$ with $\hat{\alpha}^*=\alpha^*$. The last condition should in general imply $\hat{\alpha}=\alpha$, but just now I can't reconstruct precisely why.
iI think it's best to turn this question around a bit. If such a coincidence occurs, we have two maps $X \to E$. Thus, we can consider the space of maps $X \to E$. If someone hands you a space of covers, if its dimension is larger than the space of maps of any individual element, you can conclude that a generic pair of elements are nonisomorphic.
If $g(E)\geq2$, this mapping space must be finite, because a deformation of a map $f$ would give a global section of $f^* T_E$, which is negative. By the same logic, if $g(E)=1$, the mapping space had dimension at most $1$. If $g(E)=0$, we can use Riemann-Roch to count maps. Since a map to $\mathbb P^1$ is a two-dimensional quotient of a line bundle up to scaling, the space of them has dimension at most $g(X)+2(d(f)+1-g(X))-1$.
