In [this](https://mathoverflow.net/questions/20392/conjugating-a-subgroup-of-a-group-into-a-proper-subgroup-of-itself) question, I asked whether there existed groups $G$ with finitely presentable subgroups $H$ such that $gHg^{-1}$ is a proper subgroup of $H$ for some $g \in G$. Robin Chapman pointed out that the group of affine automorphisms of $\mathbb{Q}$ contains examples where $H \cong \mathbb{Z}$.


This leads me to the following more general question. A group $\Gamma$ is "coHopfian" if any injection $\Gamma \hookrightarrow \Gamma$ is an isomorphism. To put it another way, $\Gamma$ does not contain any proper subgroup isomorphic to itself. The canonical example of a non-coHopfian group is a free group $F\_n$ on $n$ letters. Chapman's example exploits the fact that $F\_1 \cong \mathbb{Z}$ contains proper subgroups $k \mathbb{Z}$ isomorphic to $\mathbb{Z}$.


Now let $\Gamma$ be a non-coHopfian group and let $\Gamma' \subset \Gamma$ be a proper subgroup with $\Gamma' \cong \Gamma$. Question : does there exist a group $\Gamma''$ such that $\Gamma \subset \Gamma''$ and an automorphism $\phi$ of $\Gamma''$ such that $\phi(\Gamma) = \Gamma'$? How about if we restrict ourselves to the cases where $\Gamma$ and $\Gamma''$ are finitely presentable? I expect that the answer is "no", and I'd be interested in conditions that would assure that it is "yes".


If such a $\Gamma''$ existed, then we could construct an example answering my linked-to question above by taking $G$ to be the semidirect product of $\Gamma''$ and $\mathbb{Z}$ with $\mathbb{Z}$ acting on $\Gamma''$ via $\phi$. This question thus can be viewed as asking whether Chapman's answer really used something special about $\mathbb{Z}$.



Automorphisms of supergroups of non-coHopfian groups
Let $\alpha: \Gamma\to\Gamma$ be an injection sending $\Gamma$ to $\Gamma'$. Then the $\Gamma''$ you're looking for is the infinite amalgamated product


$\cdots \*\_{\Gamma}\Gamma \*\_{\Gamma}\*\Gamma\*\_{\Gamma}\cdots$


where, at each stage, $\Gamma$ maps to the left by the identity and to the right by $\alpha$. Now the 'shift' automorphism has the property that you want, and the semidirect product with $\mathbb{Z}$ that you suggest is just the ascending HNN extension of $\Gamma$ via $\alpha$.


