Is there a commutative (or even symmetric) Schur ring $S\subset\mathbb{C}G$ over a non-abelian group $G$, which is not isomorphic (preserving both the products) to a Schur ring $S'\subset\mathbb{C}G'$ over an abelian group $G'$ with $|G'|=|G|$?
Equivalently, is there a commutative (or even symmetric) translation association scheme over a non-abelian group $G$, whose intersection numbers do not coincide to any translation association scheme over any abelian group $G'$, $|G|=|G'|$?
More specifically, is there a strongly regular graph with some parameters $(n,k,\lambda,\mu)$ which is Cayley with respect to some non-abelian group $G$, but it is not Cayley with respect to any abelian group $G'$ and, moreover, no other srg with parameters $(n,k,\lambda,\mu)$ is Cayley with respect to an abelian $G'$?
What I found out: There is the database of all SRG's. You can filter the candidates as follows. If SRG with parameters $(n,k,\lambda,\mu)$ is Cayley w.r.t. some abelian group, then there must exist its dual, whose valency $k'$ must equal to the multiplicity $f$ or $g$ of some of the eigenvalues of the original graph. (By the way, this is what I am actually interested in. I want an association scheme that has no dual.) These are computed in the database, so you can filter out those that do not have such a dual. Well, but I have no idea, how can I figure out, which of them could be Cayley w.r.t. some non-abelian group.
By the way, I'd also appreciate examples of SRG's that are Cayley w.r.t. non-abelian groups even though they might as well be Cayley w.r.t. abelian groups. Examples that I figured out so far: 1) The complete bipartite graph $K_{n,n}$ is Cayley w.r.t. $\mathbb{Z}_n\times\mathbb{Z}_n$ as well as the dihedral group $D_n$. 2) If I am not mistaken, taking any group $G$, its Cayley table should be a Latin square, which induces a SRG, which will be Cayley w.r.t. $G\times G^{\rm op}$ or something.
