I suspect that the answer to my question is well-known to be no. To be more precise, let
$G$ and $H$ be nonisomorphic finite groups of the same order. Let $S \subseteq G$ and $T \subseteq H$ be subsets satisfying the three properties: (1) the subsets are symmetric, that is $S = S^{-1}$ and $T = T^{-1}$; (2) they are minimal symmetric generating sets; (3) the size of $S$ is equal to the size of $T$. Is it possible for the Cayley graph of the pair $(G,S)$ and the Cayley graph of the pair $(H,T)$ to be isomorphic?


If the answer is yes, what is the smallest such example?



Does a Cayley graph on a minimal symmetric set of generators determine a finite group up to isomorphism?
The truncated cube (polyhedron with eight triangular faces and six octagonal faces) is a Cayley graph of both the symmetric group on four items (generators: transpose first two of the four items, rotate the last three) and of a different group that acts on 3-bit binary strings (generators: rotate the string, flip its first bit). You can tell they're different Cayley graphs because the graph isomorphism does not preserve the Cayley labeling: in one of the two Cayley graphs, the generators labeling the triangles are inverted on half of the triangles compared to the labeling of the other graph. See [this blog post](http://11011110.livejournal.com/116782.html).


