= Solution
{c}
Suppose $\sigma\ne e$. Choose a symbol $i$ moved by $\sigma$ and write $j=\sigma(i)\ne i$. Since $n\geq3$, choose a third symbol $k$ distinct from $i,j$. Let $\tau=(jk)$, a <transposition> that fixes $i$. Then
$$
(\tau^{-1}\sigma\tau)(i)=\tau^{-1}(\sigma(i))=\tau^{-1}(j)=k\ne j=\sigma(i).
$$
Thus $\tau^{-1}\sigma\tau\ne\sigma$, proving that the <conjugacy class> of any nonidentity <permutation> has more than one element. By part (a)(iii), no such <permutation> lies in the <center of a group>. The identity always does, so
$$
\boxed{Z(S_n)=\{e\}\quad(n\geq3).}
$$
The restriction on $n$ is necessary: $S_2$ is an <abelian group> and equals its own center.
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