Existence of $2$ permutations of $S_n$ such that $\pi_1 \circ{} \pi_1 ^{-1}= \pi_2\circ{}\pi_2^{-1}=\pi_1 \circ{} \pi_2^{-1}=[n n-1 ... 1]$

Question

I have been thinking about this problem for a while but I can't think of any solution. I can think of a solution for $\pi_1 \circ{} \pi_1^{-1} = \pi_2\circ{}\pi_2^{-1}=\pi_1 \circ{} \pi_2^{-1}=Id$ such as any $\pi_1=\pi_2=$ an involution (for example: $\pi_1=(12)$ then $\pi_1\circ{}\pi_1^{-1}=Id$). However, I can't think of any satisfying the opposite of the Identity. Any help?

Answer 1

I just realized I missed the solution because of my incorrect notation. Let me explain. $[nn-1...1]$ is a permutation that can be represented as $$(1\text{ }n)\circ{}(2\text{ }n-1)\circ{}...\circ{}(\frac{n}{2}\text{ }\frac{n-2}{2}) \text{ if n is even}$$ $$(1\text{ }n)\circ{}(2\text{ }n-1)\circ{}...\circ{}(\frac{n-1}{2}\text{ }\frac{n-3}{2}) \text{ if n is odd}$$ For example, take $n=6$. Then the reverse permutation is $(16)\circ{(25)}\circ{}(34)$. Now I am looking for two permutations $\pi_1\circ{\pi_2^{-1}}=(16)\circ{(25)}\circ{}(34)$ or $\pi_1=(16)\circ{(25)}\circ{}(34)\circ{}\pi_2$.

So I will let $\pi_2$ be any involution, say $\pi_2=(34)$. Then $\pi_1=(16)\circ{(25)}\circ{}(34)\circ{}(34)=(16)\circ{(25)}$ which is also a involution. So $\pi_1$ and $\pi_2$ satisfy all three conditions.

Edit: My solution is incorrect. It doesn't satisfy $\pi_1\circ{}\pi_1^{-1}=\pi_2\circ{}\pi_2^{-1}=[nn-1...1]$