Consider permutations $\pi$ of the set $\{1,\dots,n\}$ having the symmetry property $\pi \pi^* \pi = \pi^*$, where $\pi^*$ is the "reflection" $k \mapsto n+1k$. Are there references or other information about such a $\pi$?

You really do mean $\pi\pi^*\pi$, not $\pi\pi^*\pi^{1}$? – LSpice Sep 12 at 19:09

2Also, is $\pi^*$ meant to depend on $\pi$, say as $k \mapsto n + 1  \pi(k)$, or is it really a fixed permutation independent of $\pi$? If the latter, then you are just looking for permutations conjugate by a fixed involution to their inverses. – LSpice Sep 12 at 19:10

4In other words, $\pi \pi^{*} $ is involution. It should reduce the questions about your permutations to the questions about involutions. – Fedor Petrov Sep 12 at 19:11

@FedorPetrov's rephrasing is better than mine. – LSpice Sep 12 at 19:11

1Why don't you use a notation without the letter $\pi$ to denote the reflection? – YCor Sep 15 at 16:59
The problem boils down to finding permutations $\pi$ such that $\pi(n+1x)=y\iff\pi(n+1y)=x$. Let $X\subset\{1,...,n\}$ be such that $n\lvert X\rvert$ is even and let $P$ be a partition of $\{1,...,n\}\backslash X$ into pairs and denote, for every element $k$ of $\{1,...,n\}\backslash X$, its companion by $P(k)$. Then, the permutation defined by $\pi(n+1k)=P(k)$ for every $k\in\{1,...,n\}\backslash X $ and $\pi(n+1x)=x$ for every $x\in X$ satisfies the required condition. It is clear that appropriate permutations are in a onetoone correspondence with $(X,P)$ couples.


I thought a constructive/descriptive approach might be helpful depending on the OP's needs indeed. I just lightened my response a bit. – Nicéphore Bayekula Sep 15 at 16:05

By the way, as a TeX matter,
\mid
is only intended as a binary operator (like2 \mid 6
: $2 \mid 6$). For delimiters, you want\lvert
and\rvert
(like\lvert2\rvert
: $\lvert2\rvert$). I made the edit here. – LSpice Sep 15 at 17:41