Suppose I have a homomorphism $\phi: {\rm dom(\phi)}\to G$, where I know ${\rm dom}(\phi)$ is a finite index subgroup of $G$. I am interested in what happens when you look at a high iterate of this homomorphism. This gets complicated though, since ${\rm dom}(\phi)$ is potententially not invariant. In other words, $\phi({\rm dom}(\phi))\not\subseteq {\rm dom}(\phi),$ so $\phi$ is not in general self-composable.
For the rest of the post, I will write $\phi^2=\phi\circ\phi$ to denote the homomorphism defined on the restricted domain ${\rm dom}(\phi)\cap \phi({\rm dom}(\phi))$, which is exactly where this homomorphism is possible to define. I will denote this restricted domain as ${\rm dom}(\phi^2)$. In general, if $K_1={\rm dom}(\phi)$, then $\phi^n$ will be used to denote the iterated homomorphism defined on $K_n = K_{n-1}\cap \phi(K_{n-1})$, and I will write $K_n=:{\rm dom}(\phi^n)$.
Now, it's not too hard to show that ${\rm dom}(\phi^N)$ will have finite index in $G$. You just show that if $[G:{\rm dom}(\phi)] =:d$, then $[{\rm dom}(\phi):{\rm dom}(\phi^2)]\leq d$ as well. Then you can use that $$G\geq {\rm dom}(\phi)\geq {\rm dom}(\phi^2)\geq \dotsm\geq{\rm dom}(\phi^N)$$ to conclude that $[G:{\rm dom}(\phi)]\leq d^N$.
My problem requires infinite iteration though. So my question is when will $$K:=\bigcap_{n=1}^\infty {\rm dom}(\phi^n)$$ be nontrivial and finite index in $G$? Although I have no explicit example, I am almost certain that the nontrivial-ness and finite-index will not be true in general. But are there nice (optimistically: easily checkable) properties that might guarantee this? It could even be looser: rather than $K$ itself being finite index, I would be happy to find nonempty $H\leq K$ with $[G:H]<\infty$.
Even if there aren't nice answers to my soft questions, does anyone know of a reference discussing "iterated homomorphisms"?
Edit: changed "nonempty" to "nontrivial"