Considering a permutation of [1, 2, ..., n], it is fairly obvious that on doing n/2 swaps we arrive at the permutation [n, n-1, ..., 1]. This can be achieved by swapping the first element with the last, the second with the second from the last, and so on.
The part I'm having trouble with is writing a formal theoretical proof for this problem. I tried my hand at induction but haven't made much progress.
Any pointers would be well appreciated !
A formal theoretical proof. Let $\sigma$ be the inverse permutation of the set $[n]=\{1,\dots,n\}$, that is $\sigma(i)=n+1-i$ for each $i\in [n]$. For each $1\le i\le \lfloor \tfrac n2\rfloor$ let $\sigma_i$ be a transposition of elements $i$ and $n+1-i$, that is $\sigma_i(i)=n+1-i$ and $\sigma_i(n+1-i)=i$. Let $\rho=\prod_{i=1}^{\left\lfloor \tfrac n2\right\rfloor}\sigma_i$ and $j\in n$ be any element. If $n$ is odd and $j=\tfrac{n+1}2$ then $\rho(j)=j=\sigma(j)$. Otherwise there exists a unique $i$, $1\le i\le \tfrac n2$ such that $j\in \{i,n+1-i\}$. Then $\rho(j)=\sigma_i(j)=n+1-j=\sigma(j).$ That is $\rho=\sigma$.