The simple random walk on the circle graph $\mathbb{Z}/N\mathbb{Z}$ of size $N$ can be realized as the class modulo $N$ of simple random walk on $\mathbb{Z}$, hence $$ p_{00}^{(2n)}(\mathbb{Z}/N\mathbb{Z})=\sum_{k\in\mathbb{Z}}p_{0,kN}^{(2n)}(\mathbb{Z}). $$ Now, for every $k$, $$ p_{0,2k}^{(2n)}(\mathbb{Z})=2^{-2n}{2n\choose n+k}, $$ with the convention that ${2n\choose i}=0$ for every $i\le -1$ or $i\ge 2n+1$.

The simple random walk on the circle graph $\mathbb{Z}/N\mathbb{Z}$ of size $N$ can be realized as the class modulo $N$ of simple random walk on $\mathbb{Z}$, hence $$ p_{00}^{(n)}(\mathbb{Z}/N\mathbb{Z})=\sum_{k\in\mathbb{Z}}p_{0,kN}^{(n)}(\mathbb{Z}). $$ Now, for every $k$, $$ p_{0,k}^{(n)}(\mathbb{Z})=2^{-n}{n\choose (n+k)/2}, $$ with the convention that ${n\choose i}=0$ for every noninteger $i$ and every integer $i\le -1$ and every integer $i\ge n+1$.

On the other hand, this is also a random walk on a finite graph hence the stationary distribution allows to shortcut these exact computations for large $n$. For every odd $N$, $$ \lim_{n\to+\infty}p_{00}^{(n)}(\mathbb{Z}/N\mathbb{Z})=\frac1N. $$ For every even $N$, $p_{00}^{(2n+1)}(\mathbb{Z}/N\mathbb{Z})=0$ for every $n$ and $$ \lim_{n\to+\infty}p_{00}^{(2n)}(\mathbb{Z}/N\mathbb{Z})=\frac2N. $$