Let us consider the Cartesian product $X^r$, where $X$ is a smooth projective variety. There is a subgroup $Aut_{\Delta}(X^r)\subset Aut(X^r)$ of automorphisms of $X^r$ mapping the small diagonal $\Delta\subset X^r$ isomorphically to itself.

If we consider the diagonal action of $Aut(X)$ on $X^r$ clearly we have $S_r\times Aut(X)\subseteq Aut_{\Delta}(X^r)$, where $S_r$ is the symmetric group. Under which hypothesis on $X$ (perhaps $X$ of general type) do we have $S_r\times Aut(X)= Aut_{\Delta}(X^r)$?