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 $$~$\sigma = (a_{11} a_{12} \dots a_{1 n_1})(a_{21} \dots a_{2 n_2}) \dots (a_{k 1} a_{k 2} \dots a_{k n_k})$~$$ Let $~$\tau \in S_n$~$.

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

Therefore $$~$\tau \sigma \tau^{-1} = (\tau(a_{11}) \tau(a_{12}) \dots \tau(a_{1 n_1}))(\tau(a_{21}) \dots \tau(a_{2 n_2})) \dots (\tau(a_{k 1}) \tau(a_{k 2}) \dots \tau(a_{k n_k}))$~$$ which has the same cycle type as $~$\sigma$~$ did.

Same cycle type implies same conjugacy class

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

Then define $~$\tau(a_{ij}) = 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$~$.