Can one distinguish finite groups by their maps from abelian groups? Given a finite group $G$, is $G$ determined by its category of maps from abelian groups? Specifically, we can form the category $G_A$ of "abelian points" of $G$ with objects pairs $(A,\phi)$, with $\phi:A\rightarrow G$, for $A$ abelian, with morphisms $\xi:(A,\phi)\rightarrow (B,\psi)$ given by morphisms of groups $\xi:B\rightarrow A$ such that $\phi \circ \xi = \psi$.
Very similar questions have been asked before, but they focused on the maps from cyclic groups, and the (small) counterexamples given for the analagous question do not have isomorphic categories of abelian points.
The following are the similar questions I refer to:
Is a finite group uniquely determined by the orders of its elements?
If I know the order of every element in a group, do I know the group?
 A: Not a full answer but hopefully a good starting point. Let $G_1,G_2$ be two groups and $\varphi\colon G_1 \to G_2$ a homomorphism. We'll say $\varphi$ is good if in the semi-direct product $G_1 \rtimes_\varphi G_2$ elements $x,y$ commute iff either $x,y\in G_1$ or $x,y\in G_2$. In that case, any homomorphism $A\to G_1\rtimes_\varphi G_2$ from an abelian group lands in either $G_1$ or $G_2$. In other words, if $\varphi$ is good, then $(G_1\rtimes_\varphi G_2)_A\cong (G_1/A)\times (G_2/A)$. In particular, if you can find two good twisting homomorphisms $\varphi_1,\varphi_2$, then the category invariant you consider will not be able to differentiate between $G_1\rtimes_{\varphi_1} G_2$ and $G_1 \rtimes_{\varphi_2}G_2$. If $\varphi_1,\varphi_2$ can be chosen to produce non-isomorpihc semi-direct products, then you're done. This latter property is easy to achieve (generally, semi-direct products on the same factors are highly sensitive to the twisting homomorphism, even for very small groups). I suspect making sure the twisting homomorphisms are also good is possible, though I don't have a construction right now. 
A: No. Consider two semidirect products $(\mathbf{Z}/7\mathbf{Z})^2\rtimes(\mathbf{Z}/3\mathbf{Z})$; where the canonical generator of $\mathbf{Z}/3\mathbf{Z}$ acts in the first case by the matrix $\begin{pmatrix}2 & 0\\0 & 2\end{pmatrix}$, and in the second case by $\begin{pmatrix}2 & 0\\0 & 4\end{pmatrix}$. They're clearly not isomorphic, but they have the same "combinatorics" of abelian subgroups: the unique $7$-Sylow subgroup (elementary abelian of order $7^2$) and its subgroups, and $7^2$ subgroups of order 3, and all these have pairwise trivial intersection.
