correspond to the same eigenvalue, denoted by $\lambda_k$. Note that $|\Upsilon_k|={n\choose k}$. It means that the multiplicity of $\lambda_k$ is ${n\choose k}$. It completes the proof.

At a glance of Theorem \[thm-z-1\], one may think that any orbit Cayley graph over $Z_2^n$ has exactly $n+1$ distinct eigenvalues. However, the values $\lambda_k$ and $\lambda_{k'}$ may be the same for distinct integers $k$ and $k'$. Thus, Theorem \[thm-z-1\] just implies the following result.

\[cor-z-2\] If $Cay(Z_2^n,S)$ is an orbit Cayley graph, then it has at most $n+1$ distinct eigenvalues.

Note that, if $S=Z_2^n\setminus \{\mathbf{0}\}$, then the orbit Cayley graph $Cay(Z_2^n,S)$ is isomorphic to the complete graph, which has exactly two distinct eigenvalues. In general, we pose the following problem.

\[conj-d-1\] If there exists an orbit Cayley graph over $Z_2^n$ with exactly $m$ distinct eigenvalues for any integer $m$ such that $2\le m\le n+1$ ?

The answer is positive for $m=2$. The following examples imply that the answer is also positive for $m=n+1$, $\lfloor n/2\rfloor+2$ and $3$, respectively. Note that, if an $r$-regular graph has eigenvalues $r>\lambda_2\ge\cdots\ge\lambda_n$, then its complement has eigenvalues $n-1-r>-1-\lambda_n>\cdots>-1-\lambda_2$ [@Cvetkovic].

\[exa-1\] We consider the orbit Cayley graph $X=Cay(Z_2^n,S)$ where $S=Z_2^n\setminus(\{\mathbf{0}\}\cup O_1)$. It is clear that $\overline{X}=Cay(Z_2^n,O_1)$. By Theorem \[thm-z-1\], 