Automorphism sending subgroup to a strict subset

Question

I am thinking about the following statement and can't seem to be able to either prove it of disprove it:

Let $G$ be a finitely generated (or even finitely presented) group, let $H < G$ be a subgroup and consider any automorphism $\varphi \in{\rm Aut}(G)$. Then it is impossible that $\varphi(H) \subsetneq H$.

I know this can easily happen if $G$ is not finitely generated.

Answer 1

It is possible. Take $H$ to be a free $2$-generated subgroup of a free group $F$ and $L$ be a free $2$-generated proper subgroup of $H$. Let $G$ be the HNN extension of $F$ with associated subgroups $H, L$. Then $H, L$ are conjugate in $G$. So there exists an inner automorphism of $G$ which takes $H$ to $L$. If $F$ is finitely generated, then $G$ is always finitely presented.

Answer 2

Let $G = \langle x,y \rangle \leq GL(2,\mathbb{Q})$, where $x = \begin{pmatrix} 1 & 1 \\\ 0 & 1 \end{pmatrix}$ and $y = \begin{pmatrix} 2 & 0 \\0 & 1 \end{pmatrix}$.

Then for $H = \langle x \rangle$ we have $yHy^{-1} = \langle x^2 \rangle \lneq H$. So for your question consider the inner automorphism $\varphi$ corresponding to $y$.

This is the example mentioned by Derek Holt in the comments.