RO-NMO 2019 Problem 114

Let $p$ be a prime number. For any $\sigma \in S_p$ (the permutation group of $\{1,2,...,p \}),$ define the matrix $A_{\sigma}=(a_{ij}) \in \mathcal{M}_p(\mathbb{Z})$ as $a_{ij} = \sigma^{i-1}(j),$ where $\sigma^0$ is the identity permutation and $\sigma^k = \underbrace{\sigma \circ \sigma \circ ... \circ \sigma}_k.$ Prove that $D = \{ |\det A_{\sigma}| : \sigma \in S_p \}$ has at most $1+ (p-2)!$ elements.