Let $G$ be a Fuchsian group, by which we mean a discrete subgroup of ${\rm PSL}(2,\mathbb{R})$. Or equivalently: $G$ is a subgroup of the automorphisms of $\mathbb{H}$ which acts properly discontinuously.
Let $\phi: G\to{\rm PSL}(2,\mathbb{R})$ be a homomorphism. When can we say $\phi(G)$ is also Fuchsian?
I believe it's not true in general, but I've been having trouble constructing a counterexample. Here's what I have so far:
Let's suppose $G$ is finitely generated with at least two generators $g_1,g_2$, and suppose further there is no relation between them. Define $\phi(g_1)$ to be $z\mapsto z+1$ and $\phi(g_2)\mapsto kz$ for $0<k<1$, and then send everything else in $G\setminus \langle g_1,g_2\rangle$ to the identity. I believe(?) this gives a well-defined homomorphism by extending in the natural way. The image $\phi(G)$ is not Fuchsian then since one can construct indiscrete orbits (for example, $\phi(G)\cdot i$ includes points of the form $i+mk^n$, and so $i$ is not isolated in the orbit).
There seems to be issues with this idea though. For one thing, one can't send the generators of $G$ to elements of arbitrary type. For example, if $g\in G$ is elliptic and of finite order, then $\phi(g)$ must be elliptic as well. It would also be nice to not have to assume $G$ is finitely generated, or make arbitrary assumptions about the generators. Does anyone have a more general construction?
A softer question: what conditions can be imposed to ensure that $\phi(G)$ is Fuchsian? Does assuming $G$ is, e.g., nonelementary and elliptic-free win you anything? Or does one need to impose restrictions on $\phi$ (say, make it continuous)?
This is not a full answer, just too long for a comment.
Any map from a Fuchsian group is continuous, as the topology on the group is discrete. Requiring that the homomorphism is a homeomorphism on its image is clearly sufficient, as being Fuchsian is a purely topological property.
Being a discrete subgroup is a strong condition, as the group structure forces the group to be uniformly discrete, i.e. there is a radius such that a ball of this radius around any point in the group intersects the group only in that point. Thus, you can reduce to trying to find some Fuchsian group that contains in its image a non-repeating sequence in $PSL_2(\Bbb{R})$ that approaches $I$. In a Fuchsian group, any sequence approaching $I$ must eventually be constant $I$, so this would show the claim.
This is difficult, and might be impossible. For example, there is no matrix $A\neq I$ such that $A^n\to I$ (this can be shown by an eigenvalue argument), so there is no hope that $\Bbb{Z}$, embedded as a Fuchsian group, could map to something non-Fuchsian.