Consider the problem:
Let $G$ be a finite group. Let $H, K$ be arbitrary subgroups of $G$. Let $f: H\to K$ be an isomorphism on groups. Can we always construct $F\in\text{Aut}\,(G)$ such that $F_{|H}=f$?
I am quite sure the answer is no. To this end, I am seeking the following:
- A concrete example of when this fails.
- A characterization of groups where this condition fails/does not fail from a representation-theoretic perspective (if it exists).
I have searched around a bit and not come up with much else other than this thread on MO and this thread on MSE from ten years ago, so links to (recent?) research papers are very much appreciated!
Edit: I am now aware of a paper by Pettet and another one by Schupp both of which assert that an automorphism $\phi: G\to G$ is extendable to an automorphism $\psi: E\to E$ where there is a natural inclusion $\iota: G\to E$ if and only if $\phi$ is inner. I suppose what I am seeking now is an equivalent formulation and proof of this problem (if one exists) from a representation-theoretic perspective. I would also be interested in approaches making use of homological algebra.
Example 1. A normal subgroup isomorphic to a nonnormal subgroup.
$H=\{(1),\ (1\ 2)(3\ 4),\ (1\ 3)(2\ 4),\ (1\ 4)(2\ 3)\}$ and $K=\{(1),\ (1\ 2),\ (3\ 4),\ (1\ 2)(3\ 4)\}$ are isomorphic subgroups of $S_4$, but $H$ is a normal subgroup of $S_4$ and $K$ is not. An automorphism of a group must take normal subgroups to normal subgroups.
Example 2. An Abelian example.
$G=\langle(1\ 2\ 3\ 4),\ (5\ 6)\rangle$ is an Abelian group of order $8$. The subgroups $H=\langle(1\ 2\ 3\ 4)^2\rangle$ and $K=\langle(5\ 6)\rangle$ are isomorphic subgroups of order $2$. No automorphism of $G$ maps $(1\ 2\ 3\ 4)^2$ to $(5\ 6)$, since the odd permutation $(5\ 6)$ is not a square; also, the quotient groups $G/H$ and $G/K$ are nonisomorphic.