Rigidity of isometries of finite covers of Riemann surfaces Let $\Theta$ and $\Sigma$ be compact Riemann surfaces with $\widetilde\Theta=\widetilde\Sigma=\mathbb{H}^2$ (so $\theta$ and $\Sigma$ are quotients of $\mathbb{H}^2$ by Fuchsian groups) such that $\rho:\Theta\rightarrow\Sigma$ is a finite cover (locally isometric not branched).  
What can we say about $\textrm{Isom}(\Theta)$ if we have a description of $\textrm{Isom}(\Sigma)$?  More specifically, is every element of $\textrm{Isom}(\Theta)$ a lift of an element of $\textrm{Isom}(\Sigma)$?
If this is too general, then assume that $\Sigma$ is constructed from a Fricke canonical polygon with $4g$ sides.  Note that $\textrm{Isom}(\Theta)$ and $\textrm{Isom}(\Sigma)$ are necessarily finite.
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As an additional restriction assume that the Fuchsian group $\Gamma$ used to construct $\Sigma$ is torsion-free and arithmetic (i.e. $\Gamma$ is commensurable with a Fuchsian group derived from a quaternion.
 A: Let $g$ denote the genus of $\Sigma$. I will say that a Riemann surface $\Sigma$  is maximal if for every holomorphic covering map $p: \Sigma'\to \Sigma$ the group of conformal automorphisms of $\Sigma'$ consists only of elements which are lifts of automorphisms of $\Sigma$ (this includes automorphisms of the covering map $p$: these are lifts of the identity map $\Sigma\to \Sigma$). 
Suppose that $g\ge 3$. Then there exists an open and dense subset $U$ of the moduli space ${\mathcal M}_g$ of genus $g$ surfaces such that every $\Sigma\in U$ is uniformized by a (torsion-free) maximal Fuchsian subgroup of $PSL(2, {\mathbb R})$. Here a Fuchsian subgroup is called maximal if it is not contained in any strictly larger Fuchsian subgroup. 
See
L. Greenberg, Maximal Fuchsian groups. Bull. Amer. Math. Soc. 69 (1963) 569–573.
and also (for further results in this direction)
L. Greenberg, Maximal groups and signatures. Discontinuous groups and Riemann surfaces (Proc. Conf., Univ. Maryland, College Park, Md., 1973), pp. 207–226. Ann. of Math. Studies, No. 79, Princeton Univ. Press, Princeton, N.J., 1974. 
Next, observe that if $F< PSL(2,{\mathbb R})$ is a maximal (torsion-free) Fuchsian subgroup then the Riemann surface $H^2/F$ is maximal. Indeed, a finite holomorphic covering map $p: \Sigma'\to \Sigma$ comes from a 
finite-index subgroup $F'< F$: $\Sigma'=H^2/F'$. The preimage of the group of  conformal automorphisms of $\Sigma'$ in $PSL(2, {\mathbb R})$ is a 
Fuchsian subgroup $F''< PSL(2,{\mathbb R})$ containing $F'$ as a finite index subgroup. Then the subgroup of P$SL(2,{\mathbb R})$ generated by $F, F''$ would be still Fuchsian, hence, equal to $F$ by the maximality assumption. 
Thus, "generic" Riemann surfaces $\Sigma$ of genus $\ge 3$ are maximal. This argument fails in the case of $g=2$ but only because every genus 2 Riemann surface is hyperelliptic, it admits a hyperelliptic involution. By digging a bit deeper in Greenberg's papers, you can see that every generic genus 2 Riemann surface is again maximal. 
I will write more on arithmetic surfaces later. 
Edit. Here are more details on the arithmetic case. I will need a definition and a theorem:
Definition. Two subgroups $H_1, H_2$ of a group $G$ are called commensurable if $H_1\cap H_2$ has finite index in both $H_1, H_2$. Given a subgroup $H< G$ the commensurator of $H$ in $G$, denoted $Comm_G(H)$, is the subset consisting of elements $g\in G$ such that the subgroups $H^g:= gHg^{-1}< G$ and $H$ are commensurable. 
It is easy to see that $Comm_G(H)$ is a subgroup of $G$. 
Theorem (Borel?). Let $G$ be a connected semisimple Lie group. Then for every arithmetic subgroup $\Gamma< G$, the commensurator $Comm_G(\Gamma)$ is dense in $G$. 
I will use this in the case $G=PSL(2, {\mathbb R})$.
You can find a proof for instance in the book by Machlachlan and Reid "The arithmetic of hyperbolic 3-manifolds". You can easily convince yourself that in the case $\Gamma=PSL(2, {\mathbb Z})$ the commensurator 
$Comm_G(\Gamma)$ contains $PSL(2, {\mathbb Q})$. 
This theorem has a much harder converse due to Margulis (that I will not need): If the commensurator of a lattice is nondiscrete then the lattice is arithmetic. 
Proposition. Every arithmetic Riemann surface $\Sigma$ is non-maximal. 
Proof. Let $F< PSL(2, {\mathbb R})$ be a Fuchsian subgroup. Then for every $g\in Comm_G(F)$ the subgroup $F^g$ normalizes a finite index subgroup $F'$ of $F$ and, hence, acts on the Riemann surface $H^2/F'$ by automorphisms. If the surface $\Sigma=H^2/F$ were maximal, all the Fuchsian groups of the form  $F^g$ would be contained in a fixed Fuchsian group $\hat{F}$ (obtained by lifting all automorphisms of $\Sigma$). However, the union
$$
\bigcup_{g\in Comm_G(F)} F^g
$$
is stable under conjugation via elements of $Comm_G(F)$. Hence, the subgroup $M$ generated by this union is also normalized by $Comm_G(F)$. In view of the density of $Comm_G(F)$ in $G$, the subgroup $M$ has to be dense as well. This means that $M$ cannot be contained in a discrete subgroup $\hat{F}$ as above, proving non-maximality of the surface $\Sigma$. qed
This proof is nonconstructive. You will obtain explicit examples by considering a non-uniform lattice in $G$ such as $F=PSL(2, {\mathbb Z})$ and looking at its conjugates $F^g$ by, say, linear-fractional maps of the form $z\mapsto az$, $a\in {\mathbb Q}^2$. 
