If the Weyl group $\mathcal W$ is a normal subgroup of $\mathrm{Aut}(\Phi)$, how can the Weyl group of $A_2$ be dihedral of order $6$?
The roots of $A_2$ are $\{\pm \alpha, \pm \beta, \pm(\alpha+\beta)\}$. So, $\mathrm{Aut}(\Phi) \cong S_6$. But, $S_6$ does not have a subgroup isomorphic to dihedral group of order $6$.
$\mathrm{Aut}(\Phi)$, for $\Phi$ a root system of type $A_2$, is not isomorphic to $S_6$; it is the group of linear automorphisms, you cannot freely permute all roots. Rather, this automorphism group is actually the symmetry group of a hexagon (the hexagon formed by the endpoints of the roots), which is the dihedral group of order $12$, which is the direct product of the group of order $2$ and the dihedral group of order $6$; this second factor is the Weyl group, and being a direct factor, it's of course normal.