Suppose there is a bijection between the underlying sets of two finite groups $G, H$, such that every subgroup of $G$ corresponds to a subgroup of $H$, and that every subgroup of $H$ corresponds to a subgroup of $G$. Does this imply that $G, H$ are isomorphic? Note that we do not require the bijection to actually be the isomorphism.
The question is interesting to me because I am considering maps of groups which aren't homomorphisms but preserve the subgroup structure in some sense - given a group, we can forget the multiplication operation and look only at the closure operator that maps a subset of $G$ to the subgroup generated by it. If the question is resolved in the affirmative, then the forgetful functor from the usual category $Grp$ to this category won't create any new isomorphisms. (Note that I didn't precisely specify the morphisms this new category -- you could just use the usual definition of a homomorphism, and say that if the mapping commutes with the closure operator, then its a morphism. The definition I actually care about is, a morphism of this category is a mapping such that every closed set in the source object is the preimage of a closed set of the target object. It doesn't make much difference as far as this question is concerned, the isomorphisms of both categories are the same.)
I asked a friend at Mathcamp about this a few weeks ago, he said a bunch of people started thinking about it but got stumped after a while. The consensus seems to have been that it is probably false, but the only counter examples may be very large. I don't really have any good ideas / tools for how to prove it might be true, I mostly wanted to just ask if anyone knew offhand / had good intuition for how to find a finite counterexample.
Edit (YCor): (a) the question has reappeared in the following formulation: does the hypergraph structure of the set of subgroups of a (finite) group determine its isomorphism type? A hypergraph is a set endowed with a set of subsets. The hypergraph of subgroups is the data of the set of subgroups, and therefore to say that groups $G,H$ have isomorphic hypergraphs of subgroups means that there's a bijection $f:G\to H$ such that for every subset $A\subset G$, $f(A)$ is a subgroup of $H$ if and only $A$ is a subgroup of $G$. Several answers, complementing the one given here, have been provided in this question.
(b) There a weaker well-studied notion for groups, namely to have isomorphic subgroup lattices. Having isomorphic hypergraphs of subgroups requires such an isomorphism to be implemented by a bijection (this is not always the case: take two groups of distinct prime order).