The problem boils down to finding permutations $\pi$ such that $\pi(n+1-x)=y\iff\pi(n+1-y)=x$. Let $X\subset\{1,...,n\}$ be such that $n-\mid X\mid$ 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 $\pi(n+1-k)=P(k)$ for every $k\in\{1,...,n\}\backslash X $ and $\pi(n+1-x)=x$ for every $x\in X$ satisfies the required condition. It is clear that appropriate permutations are in a one-to-one correspondence with $(X,P)$ couples.

The problem boils down to finding permutations $\pi$ such that $\pi(n+1-x)=y\iff\pi(n+1-y)=x$. 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 $\pi(n+1-k)=P(k)$ for every $k\in\{1,...,n\}\backslash X $ and $\pi(n+1-x)=x$ for every $x\in X$ satisfies the required condition. It is clear that appropriate permutations are in a one-to-one correspondence with $(X,P)$ couples.

As noted above, $ \pi\pi^{*}$ must be an involution. Since this transformation is already known to be a bijection, the problem boils down to finding permutations $\pi$ such that $\pi\pi^{*}(x)=y\iff\pi\pi^{*}(y)=x$, that is, $\pi(n+1-x)=y\iff\pi(n+1-y)=x$.

Now, let $X\subset\{1,...,n\}$ be such that $n-\mid X\mid$ 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 $\pi(n+1-k)=P(k)$ for every $k\in\{1,...,n\}\backslash X $ and $\pi(n+1-x)=x$ for every $x\in X$ satisfies the required condition.

It is clear that appropriate permutations are in a one-to-one correspondence with $(X,P)$ couples.

The problem boils down to finding permutations $\pi$ such that $\pi(n+1-x)=y\iff\pi(n+1-y)=x$. Let $X\subset\{1,...,n\}$ be such that $n-\mid X\mid$ 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 $\pi(n+1-k)=P(k)$ for every $k\in\{1,...,n\}\backslash X $ and $\pi(n+1-x)=x$ for every $x\in X$ satisfies the required condition. It is clear that appropriate permutations are in a one-to-one correspondence with $(X,P)$ couples.