Let $n \geq 2$ and let $\pi : \{1,\ldots, n\} \to \{1,\ldots, n\}$ be a permutation. Next, we define a new permutation $\sigma: \{0,1\}^n \to \{0,1\}^n$ as follows. For each $x = x_n x_{n-1} \ldots x_1 \in \{0, 1\}^n$, we define $\sigma(x) = x_{\pi(n)}x_{\pi(n-1)} \ldots x_{\pi(1)}$. I am interested in determining, for what values of $n$ can we find $\pi$ such that $\sigma$ is an odd permutation?
When $n=2$, we can define $\pi$ to be the permutation $(1,2)$ which gives $\sigma(00) = 00$, $\sigma(01) = 10$, $\sigma(10) = 01$ and $\sigma(11) = 11$; an odd permutation. However, I have not been able to find such a $\pi$ for larger values of $n$.
The permutation $\pi$ can be expressed as the product of transpositions $\tau_i$. Then the corresponding $\sigma_\pi$ is the product of all $\sigma_{\tau_i}$.
Let $\tau$ be one such transposition $(a,b)$. What is the effect of $\sigma_\tau$ on a number $x\in\{0,1\}^n$?
Thus $2^{n-2}$ distinct pairs $(x,x')$ are transposed, and all other numbers are left fixed. This means that $\sigma_\tau$ can be written as the product of $2^{n-2}$ distinct transpositions. Thus $\sigma_\tau$ is even if $n>2$, and the product $\sigma_\pi$ must also be even.