In this 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 $mathbb{Z}$.
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