Consider binary (0/1) sequences of length $2n$. The permutations (symmetric group $S_{2n}$) naturally acts on any sequences on length $2n$ - just permuting the symbols.
That action satisfies the condition that $x$ and $p(x)$ contain exactly the same number of $1$ (as well as $0$).
Question: consider the map "NOT" which just sends $1->0, 0->1$, I wonder is there any permutation which gives precisely that map on subset of sequences containing exactly $n$ symbols $1$ and (as well as $0$) ?
For example it is trivially true for n=1 : not: 01 -> 10 ,10->01 . That is precisely the permutation (12).