Let $a$ be an $m$-cycle. The number of conjugates of $a$ is the number of $m$-cycles, which is: $ \frac{n(n-1)\cdots(n-m+1)}{m} $
By the orbit-stabilizer theorem, the number of conjugates of $a$ is: $ \frac{\left|S_n\right|}{\left|C_{S_n}(a)\right|} = \frac{n!}{\left|C_{S_n}(a)\right|} $
Where $C_{S_n}(a)$ is the centralizer of $a$ in $S_n$. It follows that: $ \left|C_{S_n}(a)\right| = m (n-m)! \tag{1} $
Since $a$ commutes with all elements in the cyclic group $\langle a \rangle$, we have: $ \langle a \rangle \le C_{S_n}(a) \tag{2} $
For $m \in \{n-1, n\}$, $(1)$ gives $\left|C_{S_n}(a)\right| = m$. Since $\left|\langle a \rangle\right| = m$, we have equality in $(2)$.