the spectrum of $\overline{X}$ consists of $$\lambda_k(O_1)=\sum_{j=0}^1{k\choose j}{n-k\choose i-j}(-1)^j={k\choose 0}{n-k\choose 1}-{k\choose 1}{n-k\choose 0}=n-2k$$ with multiplicity ${n\choose k}$ for $0\le k\le n$. Note that $\overline{X}$ is $n$-regular. The spectrum of $X$ consists of $\mu_0=2^n-1-n$ with multiplicity $1$ and $\mu_k=2k-1-n$ with multiplicity ${n\choose k}$ for $1\le k\le n$. Therefore, $X$ and $\overline{X}$ are both connected orbit Cayley graphs having $n+1$ distinct eigenvalues.

\[exa-2\] We consider the orbit Cayley graph $X=Cay(Z_2^n,O_2)$. By Corollary \[cor-z-1\], $X$ is disconnected and thus $\overline{X}$ is connected. By Theorem \[thm-z-1\], the eigenvalues of $X$ consist of $$\lambda_k(O_2)=\sum_{j=0}^2{k\choose j}{n-k\choose i-j}(-1)^j=\frac{n(n-1)-4k(n-k)}{2}$$ with multiplicity ${n\choose k}$ for $0\le k\le n$. Therefore, the eigenvalues of $\overline{X}$ consist of $\mu_0=2^n-1-\frac{n(n-1)}{2}$ with multiplicity $1$ and $\mu_k=-1-\frac{n(n-1)-4k(n-k)}{2}$ with multiplicity ${n\choose k}$ for $1\le k\le n$. Note that $\lambda_k=-1-\frac{n(n-1)-4k(n-k)}{2}=-1-\frac{n(n-1)-4k'(n-k')}{2}=\lambda_{k'}$ if and only if $k'=k$ or $k'=n-k$. It means that the connected orbit graph $\overline{X}$ has exactly $\lfloor n/2\rfloor+2$ eigenvalues.

\[exa-3\] We consider the orbit Cayley graph $X=Cay(Z_2^n,S)$ where $S=\cup_{i=1}^{n-1}O_i$. It is clear that $\overline{X}=Cay(Z_2^n,O_n)$ which is the union of $2^{n-1}$ copies of $K_2$. Therefore, the spectrum of $\overline{X}$ consists of $\pm 1$ with multiplicity $2^{n-1}$ and thus the spectrum of $X$ consists of $2^n-2$ with multiplicity $1$, $0$ with multiplicity $2^{n-1}$ and $-2$ with multiplicity $2^{n-1}-1$. 