are adjacent and $\mu$ common neighbors whenever they are not adjacent. A strongly regular graph is said to be *trivial* if its complement is disconnected. It is clear that the complement of a trivial strongly regular graph is the union of some isomorphic complete graphs [@Godsil]. We first characterize the trivial strongly regular orbit Cayley graphs over $Z_2^n$.

\[thm-z-4\] Let $S^-=\cup_{i=1}^{n-1}O_i$ and $S^o=\cup_{i=1}^{\lceil n/2\rceil}O_{2i-1}$ be two subsets of $Z_2^n$. The orbit Cayley graph $Cay(Z_2^n,S)$ is a trivial strongly regular Cayley graph if and only if $S\in \{S^-,S^o\}$. Furthermore, $Cay(Z_2^n,S^-)$ is $(2^n, 2^n-2,2^n-4,2^n-2)$-strongly regular and $Cay(Z_2^n,S^o)$ is $(2^n,2^{n-1},0,2^{n-1})$-strongly regular.

We have proved that $Cay(Z_2^n,S^-)$ is a trivial strongly regular graph in Example \[exa-3\]. Similarly, one can verify that the complement of $Cay(Z_2^n,S^o)$ is $2K_{2^{n-1}}$ and thus it is a trivial strongly regular graph. Therefore, the sufficiency follows and we shall show the necessity in what follows. Let $X=Cay(Z_2^n,S)$ be a trivial strongly regular orbit Cayley graph. Therefore, $\overline{X}$ is disconnected and thus $\overline{S}=Z_2^{n}\setminus(\{\mathbf{0}\}\cup S)$ does not contain $O_i$ for any odd $i$ or $\overline{S}=O_n$. If the latter occurs, then we have $S=S^-$. If the former occurs, then $\overline{S}$ is the union of some $O_{2i}$. In fact, $\overline{S}$ contains all $O_{i}$ for 