Verify that for all $\t\in S(n)$: $|Z(\t)|=1^{q_1}q_1!2^{q_2}q_2!\cdots n^{q_n}q_n!$ and thus $$ |C(\t)|=\frac{n!}{1^{q_1}q_1!2^{q_2}q_2!\cdots n^{q_n}q_n!}, $$ where $q_j$ is the number of cycles of length $j$ in $\t$. In particular, if $\t$ is a cycle of length $k$, then $|Z(\t)|=k(n-k)!$.

$Z(\t)=\{\pi\in S(n):\pi\t\pi^{-1}=\t\}$ now $\pi\t\pi^{-1}$ has the same cycle structure as $\t$, i.e. it has $q_j$ cycles of length $j$ and if $\s=(n_1,\ldots,n_j)$ is a $j$-cycle in $\t$, then $\pi\s\pi^{-1}$ is the $l$-cycle $(\pi(n_1),\ldots,\pi(n_j))$ in $\pi\t\pi^{-1}$. Thus $\pi\in Z(\t)$ iff $\pi$ permutes the $q_1$ $1$-cyles, the $q_2$ $2$-cycles, etc. For the rearrangements of the $1$-cycles there are $q_1!$ possibilities. For the rearrangements of the $2$-cycles there are $q_2!$ permutations of $q_2$ $2$-cycles and for each individual $2$-cycle $(l,m)$ we have to possibilities: $(l,m)$ and $(m,l)$ - thus we get $2^{q_2}q_2!$ permutations. For the $j$-cycles we have $q_j!$ permutations of $j$-cycles and for each individual $j$-cycle $\s=(n_1,\ldots,n_j)$ we have $j$ possibilities: $(n_1,\ldots,n_j)$, $(n_2,\ldots,n_j,n_1)$, $\ldots$ and $(n_j,n_1,\ldots,n_{j-1})$.