Permutation group conjugacy proof. 
I tried this example in $S_3$ and saw that this held but I couldn't figure out how to prove it for all $S_n$. Any help would be appreciated. 

First, you should show that if $\sigma$ is a cycle, then $\alpha\sigma\alpha^{-1}$ is also a cycle of the same length. In fact, you can actually write out what this permutation has to be.
Take an arbitrary permutation $\sigma = \sigma_1 \sigma_2 \dots \sigma_n$, written in disjoint cycle notation. Then $\alpha\sigma\alpha^{-1} = \alpha \sigma_1 \sigma_2 \dots \sigma_n \alpha^{-1}$. Can you find a way to rewrite this that takes advantage of the previous fact?
