In <a href="http://mathoverflow.net/questions/20392/conjugating-a-subgroup-of-a-group-into-a-proper-subgroup-of-itself">this</a> 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 $\mathbb{Z}$.