even $i$ since otherwise the component $Cay(\langle\overline{S}\rangle,\overline{S})$ of $\overline{X}$ cannot be complete. It follows that $S=S^o$. Besides, one can easily get the parameters of $Cay(Z_2^n,S^-)$ and $Cay(Z_2^n,S^-)$ because $\overline{Cay(Z_2^n,S^-)}=2^{n-1}K_2$ and $\overline{Cay(Z_2^n,S^-)}=2K_{2^{n-1}}$.

In what follows, we consider the non-trivial strongly regular orbit Cayley graphs. At first we give a criterion for a vertex transitive graph to be strongly regular.

\[lem-z-8\] Let $\Gamma$ be a connected vertex transitive graph on $n$ vertices. Then $\Gamma$ is strongly regular if and only if the partition $\Pi:$ $V(\Gamma)=\{v\}\cup N(v)\cup N^2(v)$ is an equitable partition for some $v\in V(\Gamma)$.

It is known that if $\Gamma$ is strongly regular, then the partition $\Pi$: $V(\Gamma)=\{v\}\cup N(v)\cup N^2(v)$ is equitable for any $v\in V(\Gamma)$. The necessity follows. In what follows, we show the sufficiency. It is clear that $\Gamma$ is $r$-regular for some $r$ because it is vertex transitive. Since $\Pi$ is an equitable partition, assume the quotient matrix is $$B_{\Pi}=\left(\begin{array}{ccc}0&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right),$$ where $b_{12}+b_{13}=b_{21}+b_{22}+b_{23}=b_{31}+b_{32}+b_{33}=r$. For any two vertex $x,y$ such that $x\sim y$, there exists $\sigma\in Aut(\Gamma)$ such that $x^{\sigma}=v$ and thus $y^{\sigma}=v_1$ for some $v_1\in N(v)$. Therefore, we have $|N(x)\cap N(y)|=|(N(x)\cap N(y))^{\sigma}|=|N(x^{\sigma})\cap N(y^{\sigma})|=|N(v)\cap N(v_1)|=b_{22}$. Similarly, for any two vertices $x,y\in V(\Gamma)$ such that $x\not\sim y$, we have $|N(x)\cap N(y)|=|N(v)\cap 