$|N(v')\cap N(1)|=\mu$. Therefore, by noticing $N(1)=S$, we have $|C(v',S)|=|N(v')\cap S|=|N(v')\cap N(1)|=\mu$.

Conversely, assume that $|C(v,S)|=\lambda$ for any $v\in S$ and $|C(v',S)|=\mu$ for any $v'\in \overline{S}$. It is clear that $X=Cay(G,S)$ is vertex transitive and $\Pi$: $V(X)=\{1\}\cup S\cup \overline{S}$ is a partition. Moreover, we have $|N(v)\cap S|=|C(v,S)|=\lambda$ for any $v\in S$ and $|N(v')\cap S|=|C(v',S)|=\mu$ for any $v'\in \overline{S}$. Therefore, we see that $\Pi$ is an equitable partition with quotient matrix $$B_{\Pi}=\left(\begin{array}{ccc}0&|S|&0\\1&\lambda&|S|-1-\lambda\\0&\mu&|S|-\mu\end{array}\right)\begin{array}{c}\{1\}\\S\\\overline{S}\end{array}.$$ Thus, Lemma \[lem-z-8\] implies that $X$ is $(|G|,|S|,\lambda,\mu)$-strongly regular.

Easily to show that a vertex-transitive graph with diameter two is strongly regular if it is arc-transitive. Lemma \[lem-z-8\] gives a sufficient and necessary condition for a vertex transitive graph to be strongly regular, and Theorem \[cor-z-3\] also gives a sufficient and necessary condition for a Cayley graph to be strongly regular. Such a sufficient and necessary condition is essentially to verify wether $V(X)=\{1\}\cup N(1)\cup N^2(1)$ is an equitable partition, which is really weaker than the arc-transitivity.

Now we focus on the orbit Cayley graphs over $Z_2^n$. Recall that $O_i$ are the orbits of $Z_2^{n}$ under the action of $W$ for $0\le i\le n$. Denote by $S_0=\cup_{i\ne 0,i\equiv 0(\mod 4)}O_i$, $S_1=\cup_{i\equiv 1(\mod 4)}O_i$, $S_2=\cup_{i\equiv 2(\mod 4)}O_i$ and $S_3=\cup_{i\equiv 3(\mod 4)}O_i$.

\[lem-z-9\] 