I have the following general question: Given two finite groups $N$ and $H$, how can we find, using GAP, all the groups $G$ (up to an isomorphism, of course) such that $$1 \rightarrow N \rightarrow G \rightarrow H \rightarrow 1$$ is a short exact sequence? For splitting sequences, I know how to solve the problem (computing semidirect products), but I have no idea how to construct non-splitting sequences.
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Your comments indicate that you are interested in a case of $\gcd(|N|,|H|)=1$. In this situation (due to the Schur/Zassenhaus theorem) any extension is a semidirect product.
You can classify such extensions by computing the AutomorphismGroup of $N$ and computing (classes of) homomorphisms from $H$ to $\mbox{Aut}(N)$: E.g.
gap> N:=AbelianGroup([5,5,5]);
<pc group of size 125 with 3 generators>
gap> H:=SymmetricGroup(4);;
gap> au:=AutomorphismGroup(N);
<group with 4 generators>
gap> homs:=AllHomomorphismClasses(H,au);
[ [ (1,3,2), (3,4) ] -> [ [ f1, f2, f3 ] -> [ f1, f2, f3 ],
[ f1, f2, f3 ] -> [ f1, f2, f3 ] ], [...]o
gap> ext:=List(homs,x->SemidirectProduct(H,x,N));
[ <pc group of size 3000 with 7 generators>,
<pc group of size 3000 with 7 generators>,
<pc group of size 3000 with 7 generators>,
<pc group of size 3000 with 7 generators>,
<pc group of size 3000 with 7 generators>,
<pc group of size 3000 with 7 generators>,
<pc group of size 3000 with 7 generators>,
<pc group of size 3000 with 7 generators> ]
gap> List(ext,x->Length(ConjugacyClasses(x))); # show they are non-isomorphic
[ 625, 253, 325, 265, 38, 165, 26, 105 ]
TwoCohomologyGeneric) for doing exactly this (for elementary abelian $N$). At this point you would have to install the development version from github, it is not really possible to extract the code for an older release. If you have a single example let me know and I can run the code for you. $\endgroup$ – ahulpke Sep 16 at 21:35