In the comments to the question, I notice something which might be an error, or at least is an incomplete response. It is pointed out in the comments that there exist nonisomorphic groups with isomorphic subgroup lattices. While true, that fact doesn't answer this question, since it is possible to have isomorphic subgroup lattices and nonisomorphic subgroup hypergraphs. Even if you assume that the groups have the same order and isomorphic subgroup lattices, it does not immediately follow that they have isomorphic subgroup hypergraphs. What is needed is a subgroup preserving and reflecting bijection between the groups.

I am sure the negative answer to this question can be found in Roland Schmidt's book. But I would like to point to a theorem I coauthored after Schmidt's book was published, which applies to this question. Namely:

**Thm.** For any finite $N$, there is a finite set $X=X_N$ and $N$ binary operations $\circ_i$ defined on $X$ such that

(1) $G_i = (X,\circ_i)$ is a group for all $i$,

(2) $G_i\not\cong G_j$ when $i\neq j$, and

(3) for all $i, j$, the groups
$G_i^{\kappa}$ and $G_j^{\kappa}$ have exactly the same subgroups (as sets) for all cardinals $\kappa$.

The last item means that, for any fixed $\kappa$, the subgroup hypergraphs of $G_i^{\kappa}$ and $G_j^{\kappa}$ are equal for any $i$ and $j$.

The paper is

Keith A. Kearnes and Agnes Szendrei,

Groups with identical subgroup lattices in all powers.

J. Group Theory 7 (2004), no. 3, 385--402.

if two groups $G$ and $H$ both have the same number of elements of the same order, for every order, and $G$ is solvable, then so is $H$.(This is now proved I believe.) I wonder if one could ask something similar here:Suppose $G$ and $H$ have the same subgroup hypergraph and $G$ is solvable. Then $H$ is solvable... $\endgroup$4more comments