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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).

Alexander Chervov
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    Suppose such a permutation exists and among others maps $1$ to $k$. Then any sequence with $x_1=x_k$ will not be mapped to its NOT-ed value. As such sequences exist for every $n\ge2$, your example for $n=1$ is extremely unique. – Hagen von Eitzen Feb 03 '24 at 20:26
  • @HagenvonEitzen thanks ! please consider converting into answer. Please also take a look at follow-up: https://math.stackexchange.com/q/4856288/21498 – Alexander Chervov Feb 03 '24 at 20:55

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