Let $G$ a group, $p$ a prime and $X=G^p$. Let $\sigma\in S_X$ act as follows: $\sigma(x_1,...,x_p) = (x_2,...,x_p,x_1)$. Let $Y$ the subset of elements in $X$ such that $x_1x_2...x_p=1$ and let $Y^\sigma = \{y \in Y : \sigma(y) =y\}$. Prove that $|Y| = |Y^\sigma|(\bmod p)$.
I think that $Y^\sigma$ is the stabilizer but I cannot see how to proceed after that. I think this must be straightforward and relatively easy but I am just failing to see the immediate nature of the solution.