Extensions isomorphic as groups but not congruent or pseudo-congruent

Question

I'm looking for an example of a finite abelian group A and a finite group G acting trivially on A such that there are two extensions $E_1$ and $E_2$ with base A and quotient G (i.e., they are both central extensions, and hence both give corresponding elements of $H^2(G,A)$) and:

1. $E_1$ and $E_2$ are isomorphic as abstract groups.
2. Under the natural action of $\operatorname{Aut}(G) \times \operatorname{Aut}(A)$ on $H^2(G,A)$ (by pre- and post-composition with 2-cocycles that then descends to action on cohomology classes), the cohomology classes corresponding to $E_1$ and $E_2$ are not in the same orbit.

Basically condition (2) states that $E_1$ and $E_2$ are not only not congruent extensions, they are not even congruent up to a relabeling of the subgroup A and the quotient G. Another way of putting this is that there is no isomorphism between $E_1$ and $E_2$ that sends the A inside $E_1$ to the A inside $E_2$.

The analogous statement with a nontrivial action of G on A is also of interest to me. In this latter case, though, the entire group $\operatorname{Aut}(G) \times \operatorname{Aut}(A)$ does not act.

I think that examples exist (because of my experience with finding examples for similar specifications) but there may well be a proof to the contrary.

Answer 1

E = SmallGroup(32,28) is the first example. It has two central subgroups A1 and A2 isomorphic to A ≅ 2 with quotient isomorphic to SmallGroup(16,11), but A1 and A2 are not conjugate in Aut(E).

Examples such as this are reasonably common in p-groups.

Edit: You can even have such an example with G abelian: G = 4×2, A = 4×2, E = SmallGroup(64,3) = 8⋉8, Z(E) = 4×4. E has two central copies of A=4×2 that are not conjugate in Aut(G), but the quotients are both abelian and isomorphic to G.

Edit: Vipul notes you can even have E abelian of order p⁷.