Groups containing each other as finite index subgroups

Simple counterexample: $G$ is the square of an infinite dihedral group, consisting of symmetries of the ${\bf Z}^2$ lattice of the form $(x,y) \mapsto (\pm x + a, \pm y + b)$ with $a,b \in \bf Z$; and $H$ is the index-$2$ subgroup where $a \equiv b \bmod 2$. Then $H$ has index-$2$ subgroup consisting of the symmetries $(x,y) \mapsto (\pm x + 2a, \pm y + 2b)$, and this subgroup is isomorphic with $G$, but $G \not \cong H$.

The answer is "no". It is proved in https://arxiv.org/abs/0711.1014 for R. Thompson group $F$. That group contains a finite index subgroup $H$ which is not isomorphic to $F$ but contains a copy of $F$ as a subgroup of finite index.