Let $S$ be $D_n(q)$ where $q$ is a prime power. We know that a diagonal automorphism $\phi_h$ of $S$ is of the form $g\mapsto hgh^{-1}$, where $h\in \hat H$ and $\hat H$ normalizes $S$. Note that $\hat H:=\{h(\chi)\mid \chi: \mathbb Z \Phi \rightarrow \mathbb F_q^* \}$, where $\Phi$ is the corresponding root system and $\mathbb Z \Phi$ is the root lattice and $\chi$ is a group homomorphism from free abelian group $\mathbb Z\Phi$ to cyclic group $\mathbb F_q^*$.
Let $\mathrm{Irr}(S)$ be the set of irreducible characters over complex field $\mathbb C$, and let $S\leq D\leq \mathrm{Aut}(S)$ so that $D$ is generated by $S$ and all diagonal automorphisms of $S$.
Question: Is there a $\chi\in \mathrm{Irr}(S)$ so that the inertia subgroup $I_D(\chi):=\{x\in D:\chi^x=\chi\}$ is equal to $S$?
Here is a theorem of Lusztig.
Let $S$ be a finite simple group of Lie type. Then any unipotent character $ρ$ of $S$ has an extension $\widetilde{ρ}$ to the group G of inner-diagonal automorphism of $S$ such that $ρ$, $\widetilde{ρ}$ have the same inertia group in $\mathrm{Aut}(S)$. (See Propostion 2.1 Malle, G. (2008). Extensions of unipotent characters and the inductive McKay condition. Journal of Algebra, 320(7), 2963-2980.)
Based on this theorem, the $\chi$ I mentioned in Question couldn't be an unipotent character. I am curious that whether my question is true or not.
Thank you for your kind help.
