Given any cycle $a =(a_1a_2\cdots a_k)$ in $S_n$ and any element $\sigma$ in $S_n$, conjugating $a$ by $\sigma$ gives $(\sigma(a_1)\sigma(a_2)\cdots\sigma(a_k))$. Now, since $S_n$ acts $n$-transitively on the $n$ elements $a_1,\dots,a_n$ we can pick $\sigma$ such that we get any other cycle of the same length as $a$ by conjugating by $\sigma$. Since any element can be written uniquely as a product of disjoint cycles and those cycles commute, we now get that any two cycles of the same type are conjugate. On the other hand, conjugation preserves the cycle type by the same argument, so this establishes the desired bijection.