Extending an isomorphism of subgroups to an endomorphism Let $G$ be a finite group. Let $i_1:H_1\to G$ and $i_2:H_2\to G$ be two injective group homomorphisms. Let $\psi_1:H_1\to H_2$ and $\psi_2:H_2\to H_1$ be two group homomorphisms such that $\psi_1\circ \psi_2=\mathrm{Id}_{H_2}$ and $\psi_2 \circ \psi_1=\mathrm{Id}_{H_1}$. Is it possible that there is no group homomorphism $\phi:G\to G$ such that either $\phi \circ i_1=i_2\circ \psi_1$ or $\phi\circ i_2=i_1 \circ \psi_2$?
I have verified that it is not possible for groups of order at most $7$ (in the case of $S_3$ one needs to note that the automorphism group acts transitively on the set of elements of order exactly $2$).
 A: The answer is that it is indeed possible that no such extension exists, and an example arises in groups of order $8$ (but none smaller) as mentioned by Matthias Klupsch. 
Let $G=\langle r,s\mid r^4 = s^2 =1, sr=r^{-1}s\rangle$ be the dihedral group of order $8$. Then $H_1 = \{e,r^2\}$ is the center, so any automorphism of $G$ maps $H_1$ to itself. On the other hand, $H_2=\{e,s\}$ is isomorphic to $H_1$, but not central.
However, in the situation you describe, one can always find an overgroup $\mathfrak{G}$ that contains $G$, where $H_1$ and $H_2$ are not just conjugate, but the conjugation action realizes the abstract isomorphism. That is, if $G$ is a group, $H_1$ and $H_2$ are subgroups, and $\psi\colon H_1\to H_2$ is an isomorphism, then there exists a group $\mathfrak{G}$ containing $G$, and an element $t\in \mathfrak{G}$ such that for all $h\in H_1$, $tht^{-1}=\psi(h)$, so that conjugation by $t$ maps $H_1$ to $H_2$ as $\psi$. This is called an HNN Extensions (named after Higman, Bernhard Neumann, and Hanna Neumann, who first gave the construction). 
