Interplay between automorphisms of a group and an equivalence relation on subnormal subgroups

Question

I have a question that must be simple but that is giving me some trouble.

Let $H$ be a subnormal subgroup of an infinite group $G$. We say that a subnormal subgroup (of the same group) $D$ is a near complement to $H$ in $G$ if the following hold:

(1) $\langle H,D\rangle=H\times D$

(2) $|G/(H\times D)|$ is finite

Now, we define an equivalence relation between subnormal subgroups of $G$ saying that they are equivalent if and only if they have the same near complement.

We arrive at the central question: given an automorphism $\phi\in\text{Aut}(G)$, is it true that $\phi(H)$ and $H$ belong to the same equivalence class? The answer seems quite immediate to me, since $H\times D\cong \phi(H)\times D$ and since $\langle H,D\rangle\cong\langle \phi(H),D\rangle$, but in the framework of the topic in which I'm working it could be quite strange.

Thanks!

Answer 1

Counterexample: take $G = \Bbb Z \times \Bbb Z$, $H = \langle (1, 0) \rangle$, $D = \langle(0, 1)\rangle$, and $\phi((a, b)) = (b, a)$.