Does a Cayley graph on a minimal symmetric set of generators determine a finite group up to isomorphism?

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?

Let $G = Z_4$ be the cyclic group on 4 elements, generated by $S = \{-1,1 \}$, let $H = Z_2 \times Z_2$ be the Klein four group, generated by $T = \{(0,1),(1,0)\}$. Then $|S| = |T|$ and both Cayley graphs are isomorphic to $C_4$, the cycle of length 4.
For $n > 2$ each even cycle $C_{2n}$ is a Cayley graph for the cyclic group $Z_{2n}$ and for the dihedral group $D_n$ of order $2n$.
Another well-known example is the graph of a cube $Q_3$ which is a Cayley graph for the abelian group $Z_4 \times Z_2$ and for the dihedral group $D_4$. In the previous example the dihedral group was generated by two involutions, while in the latter case it is generated by an involution and an element of order 4.