In the symmetric group on a finite set, the conjugacy class of an element is determined exactly by its cycle type.

More precisely, two elements of $S_n$ are conjugate in $S_n$ if and only if they have the same cycle type.

Proof

Same conjugacy class implies same cycle type

Suppose $\sigma$ has the cycle type $n_1,\dots ,n_k$; write $$\sigma =(a_{11}{a}_{12}\dots a_{1n_1})(a_{21}\dots a_{2n_2})\dots (a_{k1}{a}_{k2}\dots a_{k{n}_k})$$ Let $\tau \in S_n$.

Then $\tau \sigma {\tau }^{-1}\left(\tau \right(a_{ij}\left)\right)=\tau \sigma \left(a_{ij}\right)=a_{i(j+1)}$, where $a_{i(n_i+1)}$ is taken to be $a_{i1}$.

Therefore $$\tau \sigma {\tau }^{-1}=\left(\tau \right(a_{11}\left)\tau \right(a_{12})\dots \tau (a_{1n_1}\left)\right)\left(\tau \right(a_{21})\dots \tau (a_{2n_2}\left)\right)\dots \left(\tau \right(a_{k1}\left)\tau \right(a_{k2})\dots \tau (a_{k{n}_k}\left)\right)$$ which has the same cycle type as $\sigma$ did.

Same cycle type implies same conjugacy class

Suppose $$\pi =(b_{11}{b}_{12}\dots b_{1n_1})(b_{21}\dots b_{2n_2})\dots (b_{k1}{b}_{k2}\dots b_{k{n}_k})$$ so that $\pi$ has the same cycle type as the $\sigma$ from the previous direction of the proof.

Then define $\tau \left(a_{ij}\right)=b_{ij}$, and insist that $\tau$ does not move any other elements.

Now $\tau \sigma {\tau }^{-1}=\pi$ by the final displayed equation of the previous direction of the proof, so $\sigma$ and $\pi$ are conjugate.

Example

This result makes it rather easy to list the conjugacy classes of $S_5$.