# Doubt regarding conjugations of permutations

I was studying the Barrington Theorem in https://homes.cs.washington.edu/~anuprao/pubs/CSE531Sp2020/lecture2.pdf when I found a doubt regarding permutations. Particularly, with conjugations.

It is said, in the 2nd page, 4th point, that two cyclic permutations $$\pi$$ and $$\sigma$$, always have a permutation $$\tau$$ s.t. $$\tau\pi\tau^{-1} = \sigma$$.

I have never studied permutations, and wanted to see why this is true.

The unique thing I have came up is that: $$\tau\pi\tau^{-1} = \sigma$$ iff $$\tau\pi = \sigma\tau$$ but I am stack here.

Any clue would be really appreciated.

Separate the domains of the permutation and the conjugation. If X is a set and $$\sigma$$ is some permutation of X (take $$X=\{1,2,\ldots, n\}$$, say), imagine there's a new set of Z of the same cardinality as X and a one-to-one, onto mapping $$f: Z\to X$$. Consider $$f^{-1} \sigma f$$. It's a function on Z that first maps everything to X, permutes according to $$\sigma$$, and maps back along the same "mapping lines" as f. Intuitively, "the result does to Z exactly what $$\sigma$$ does to X". Working out a few small examples should help.
But be careful: The permutations are conjugate in $$S_n$$ if they have the same cycle structure. This may not be true in subgroups of $$S_n$$. Consider $$S_4$$. The elements (1 2 3) and (1 3 2) of $$A_4$$ have the same cycle structure, but they are not conjugate in $$A_4$$.
As pointed by @D.W., Theorem 2 from the link https://www.planetmath.org/conjugacyclassesinthesymmetricgroupsn has a proof for the existance of such $$\tau$$.
In essence, consider two cycles $$\pi$$ = ($$a_1$$ ... $$a_n$$) and $$\sigma$$ = ($$b_1$$ ... $$b_n$$). We can build $$\tau$$ as the permutation that moves each $$b_i$$ to $$a_i$$.
Indeed, we see that, $$\tau^{-1} \pi \tau(b_i) = \tau^{-1} \pi (a_i) = \tau^{-1}(a_{i+1}) = b_{i+1}$$. That is, each $$b_i$$ is mapped to $$b_{i+1}$$ as $$\sigma$$ does.