I am not sure if this question actually makes sense, so any corrections/suggestions would be greatly appreciated!
Let $K$ be a polycyclic group, and let $\phi_1$ and $\phi_2$ be two automorphisms in $\operatorname{Aut}(K)$ such that $\phi_1$ has infinite order and $\phi_1\phi_2$ = $\phi_2\phi_1$. Let $\phi_1'$ be the induced automorphisms on $K \rtimes_{\phi_2}\mathbb{Z} $, where $\phi_1'$ acts trivially on $\mathbb{Z}$ component. Similarly, let $\phi_2'$ be the induced automorphisms on $K \rtimes_{\phi_1}\mathbb{Z} $, where $\phi_1'$ acts trivially on $\mathbb{Z}$ component.
Now suppose $$ (K \rtimes_{\phi_1} \mathbb{Z})\rtimes_{\phi_2'}\mathbb{Z} \cong (K \rtimes_{\phi_2} \mathbb{Z})\rtimes_{\phi_1'}\mathbb{Z}. $$
Question:Given a subgroup $H$ of $K \rtimes_{\phi_1} \mathbb{Z}$ with finite index $m$, is there a "natural" subgroup $H'$ in $(K \rtimes_{\phi_2} \mathbb{Z})\rtimes_{\phi_1'}\mathbb{Z}$ that contains the subgroup $ \{(h_1,0,h_2) : (h_1,h_2) \in H \}$ with index roughly $m$ (a multiple of "m") ?
I was thinking to take $H' = \{(h_1,n\mathbb{Z},h_2) : (h_1,h_2) \in H \}$ for some $n \in \mathbb{Z}$, unfortunately this subset is not closed in general.