I've recently encountered the following problem. Given a group $G$, a subgroup $H$ and a sequence $g_n\in G$, let $$ \liminf_{j\to\infty}H^{g_j} :=\bigcup_{n\ge 1} \bigcap_{j\ge n} H^{g_j}.$$ Here $$ H^g=g^{-1}Hg$$ denotes conjugation by the element $g$. The question is whether or not the above subgroup, denote it by $H_\infty$, is conjugated to a subgroup of $H$, i.e. if $$H_\infty\le H^g$$ for some $g\in G$.

This can quite easily be seen to be true for Noetherian groups $G$: the subgroups $$H_n:= \bigcap_{j\ge n} H^{g_j}$$ form an ascending chain of subgroups of $G$, therefore eventually becoming constant ($=H_\infty$). Since each $H_n$ satisfies $H_n\le H^{g_n}$ we find that for large enough $n$, $H_n=H_\infty\le H^{g_n}$.

I'm however interested in hyperbolic groups, where my understanding is that the Noetherian property is rather rare (I'm not really in group theory so I might be wrong) and so I'm wondering under what sort of hypotheses can we find a positive answer to the question (for an arbitrary sequence $(g_n)$ and subgroup $H\le G$. If needed one can assume $H$ to be finitely generated.)

I appreciate any help, thanks!