There is an equivalence relation between inclusion of finite groups coming from the world of subfactors:
Definition: $(H_{1} \subset G_{1}) \sim(H_{2} \subset G_{2})$ if $(R^{G_{1}} \subset R^{H_{1}})\cong(R^{G_{2}} \subset R^{H_{2}})$ as subfactors.
Here, $R$ is the hyperfinite $II_1$ factor (a particular von Neumann algebra), and the groups $G_1$ and $G_2$ act by outer automorphisms.
The notation $R^G$ refers to the fixed-point algebra.
Theorem: Let $(H \subset G)$ be a subgroup and let $K$ be a normal subgroup of $G$, contained in $H$, then:
$(H \subset G) \sim (H/K \subset G/K)$. In particular, if $H$ is itself normal: $(H \subset G) \sim (\{1\} \subset G/K) $
Theorem : $(\{1\} \subset G_{1}) \sim(\{1\} \subset G_{2})$ iff $G_1 \simeq G_2$ as groups.
Remark : the relation $\sim$ remembers the groups, but not necessarily the subgroups:
Exemple (Kodiyalam-Sunder p47) : $(\langle (1234) \rangle \subset S_4) \sim (\langle (13),(24) \rangle \subset S_4)$
Is there a purely group-theoretic reformulation of the relation $\sim$ ?
Motivations: See here and here.
Some definitions: A subfactor is an inclusion of factors. A factor is a von Neumann algebra with a trivial center. The center is the intersection with the commutant. A von Neumann algebra is an algebra of bounded operators on an Hilbert space, closed by taking bicommutant and dual. Here, $R$ is the hyperfinite $II_{1}$ factor. $R^{G}$ is the subfactor of $R$ containing all the elements of $R$ invariant under the natural action of the finite group $G$. In its thesis, Vaughan Jones shows that, for all finite group $G$, this action exists and is unique (up to outer conjugacy, see here p8), and the subfactor $R^{G} \subset R$ completely characterizes the group $G$. See the book Introduction to subfactors (1997) by Jones-Sunder.
Best Answer
For finite groups, the answer was given by Izumi in his paper "Characterization of isomorphic group-subgroup subfactors" (MR1920326). There he looks at the crossed product subfactor, but you can always take duals.
Edit after @Andre's comment:
The actual condition between the two pairs of subgroups is quite technical, and it would basically require reproducing an entire page of a 10 page article. Here is a link to the article: http://imrn.oxfordjournals.org/content/2002/34/1791.short
See also this video (27:30) of a talk of M. Izumi on this subject, at the Sunder Fest 2012.