This question might be duplicate because of a representation theory question. I don't know representation theory enough so I didn't tried to check that section. Please notify.
I heard and experienced that if $s_1,s_2 \in Sym(n)$, then there exists a $k \in Sym(n)$ such that $k^{-1}s_1k = s_2 \iff s_1$ and $s_2$ has same cycle type.
I tried to proof this statement but in the middle of it someone told me it has really short proof so I started to trying to find it but I couldn't. I hope you can help me and find that short proof.
Thanks for any help.
Best Answer
As is discussed in the comments, the following fact should be very helpful:
If $\sigma,\tau\in S_n$, with $\tau=(a_1\ldots a_m)$ an $m$-cycle, then we have the formula:
$$\sigma\circ\tau\circ\sigma^{-1}=(\sigma(a_1)\ldots\sigma(a_n))$$
For the proof, see what $\sigma\circ\tau\circ\sigma^{-1}$ does to symbols of the form $\sigma (a_i)$, and also symbols of the form $\sigma (b)$ where $b\ne a_i$.
Next, use the above formula on the cycle decomposition of a generic element $g=\tau_1\circ\ldots\circ\tau_k$ to see that conjugation by $\sigma$ preserves the cycle structure.
The other direction of the proof will also make use of the conjugation formula.