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 $$\pi \left({i}\right)$$.

Proof
Consider the effect of $$\pi \rho \pi^{-1}$$ on $$\pi \left({i}\right)$$:

$$\pi \rho \pi^{-1} \left({\pi \left({i}\right)}\right) = \pi \left({\rho \left({i}\right)}\right)$$

That is, $$\pi \rho \pi^{-1}$$ maps $$\pi \left({i}\right)$$ to $$\pi \left({\rho \left({i}\right)}\right)$$.

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