Show $\sigma^{-1} (i j)\sigma = ((i)\sigma (j)\sigma)$ 
Let $n \geq 2$ be an integer and $i, j \in \{1, 2, ..., n\} $ be distinct elements. Let $\sigma \in S_n$, Show that $\sigma^{-1} (i j)\sigma = ((i)\sigma (j)\sigma)$


let  $\tau=\sigma^{-1}(ij)\sigma$, then $((i)\sigma) \tau=((i)\sigma)\sigma^{-1} (i j)\sigma=i(ij)\sigma=e(j)\sigma=(j)\sigma$, similarly, we we can get $((j)\sigma)\tau=(i)\sigma$.
Let $k\in \{1,2,3\dots,n\}$ and $k\neq (i)\sigma\neq (j)\sigma$, then $(k)\tau =k$

I am still don't what I need to show to complete the proof, can anyone give me hit? Thanks

$$((i)\sigma)\sigma^{-1}(ij)\sigma=(i)(ij)\sigma=(j)\sigma$$
and
$$((j)\sigma)\sigma^{-1}(ij)\sigma=(j)(ij)\sigma=(j)\sigma$$
So, $(i)\sigma\mapsto(j)\sigma$ and $(j)\sigma\mapsto(i)\sigma$.
