Cycle Decomposition of Conjugate

Theorem
Let $S_n$ denote the symmetric group on $n$ letters.

Let $\pi, \rho \in S_n$.

The cycle decomposition of the permutation $\pi \rho \pi^{-1}$ can be obtained from that of $\rho$ by replacing each $i$ in the cycle decomposition of $\rho$ with $\map \pi i$.

Proof
Consider the effect of $\pi \rho \pi^{-1}$ on $\map \pi i$:


 * $\map {\pi \rho \pi^{-1} } {\map \pi i} = \map \pi {\map \rho i}$

That is:
 * $\pi \rho \pi^{-1}$ maps $\map \pi i$ to $\map \pi {\map \rho i}$

In the cycle decomposition of $\pi \rho \pi^{-1}$, $\map \pi i$ lies to the left of $\map \pi {\map \rho i}$, whereas in the cycle decomposition of $\rho$, $i$ lies to the left of $\map \rho i$.