Complete and composition factor-equivalent not implies isomorphic

Statement
It is possible to have two finite groups $$G_1$$ and $$G_2$$, both of which are fact about::complete groups, such that they are fact about::composition factor-equivalent groups: they have the same list of composition factors (i.e., every simple group occurs the same number of times in the composition series of both groups), and such that $$G_1$$ and $$G_2$$ are not isomorphic.

Facts used

 * 1) uses::Semidirect product with self-normalizing subgroup of automorphism group of coprime order implies every automorphism is inner

Construction
The construction is as follows.


 * 1) The group $$PGL(2,11)$$ has two non-isomorphic $$\{2, 3 \}$$-Hall subgroups, one isomorphic to the symmetric group on four letters and the other isomorphic to a dihedral group of order $$24$$. Both are self-normalizing subgroups inside $$PGL(2,11)$$.
 * 2) The inverse images of these under the natural projection mapping $$GL(2,11) \to PGL(2,11)$$ are two non-isomorphic subgroups of $$GL(2,11)$$ of order $$240$$. Note that they are both solvable, since their quotient by a central subgroup (the center of $$GL(2,11)$$) are the solvable groups mentioned above. Call these groups $$H_1$$ and $$H_2$$. Moreover, from step (1), both $$H_1$$ and $$H_2$$ are self-normalizing subgroups of $$GL(2,11)$$. Further, they are non-isomorphic. (For instance, $$H_1$$ has no element whose order is a multiple of $$12$$, but $$H_2$$ does).
 * 3) Let $$T$$ be the elementary Abelian group of order $$121$$ and let $$G_1 = T \rtimes H_1$$ and $$G_2 = T \rtimes H_2$$ (note that both these groups act naturally on $$T$$ since they're subgroups of $$GL(2,11)$$, its automorphism group).
 * 4) Since $$T$$ is Abelian, and both $$H_1$$ and $$H_2$$ are solvable, both $$G_1$$ and $$G_2$$ are solvable groups. Moreover, since $$H_1$$ and $$H_2$$ have the same order, $$G_1$$ and $$G_2$$ are solvable groups of the same order, hence they are composition factor-equivalent.
 * 5) By fact (1), both $$G_1$$ and $$G_2$$ are groups for which every automorphism is inner.
 * 6) An easy check shows that both $$G_1$$ and $$G_2$$ are centerless. The fact that there are no central elements inside $$T$$ follows from the fact that $$T$$ has no global fixed points under the action of either $$H_i$$. The fact that there are no central elements outside $$T$$ follows from the fact that since $$T$$ is Abelian, any element outside $$T$$ acts on $$T$$ by conjugation like some non-identity automorphism in $$H_i$$.
 * 7) $$G_1$$ and $$G_2$$ are not isomorphic: The subgroup $$T$$ is a normal Sylow subgroup in each $$G_i$$, hence it is the only subgroup of its order in each $$G_i$$. Hence, any isomorphism from $$G_1$$ to $$G_2$$ must preserve $$T$$, and yield an isomorphism $$G_1/T \cong G_2/T$$. But we know that $$H_1$$ and $$H_2$$, which are isomorphic to the respective quotients, are not isomorphic to each other.
 * 8) Combining steps (4)-(7), $$G_1$$ and $$G_2$$ are composition factor-equivalent complete groups that are not isomorphic.

GAP implementation
To make things a bit easier, first define the IsCompleteGroup function (either using the interactive interface or in a file to be included. Once this is done, the following GAP implementation constructs the example:

gap> T := ElementaryAbelianGroup(121);  gap> A := AutomorphismGroup(T);  gap> HallList := Filtered(List(ConjugacyClassesSubgroups(A),Representative),K -> Order(K) = 240); [ ,  ] gap> H1 := HallList[1];  gap> H2 := HallList[2];  gap> G1 := SemidirectProduct(H1,T);  gap> G2 := SemidirectProduct(H2,T);  gap> Order(G1) = Order(G2); true gap> IsSolvableGroup(G1); true gap> IsSolvableGroup(G2); true gap> IsCompleteGroup(G1); true gap> IsCompleteGroup(G2); true gap> IsomorphismGroups(G1,G2); fail