Let $G$ be a Chevalley group over a field $k$ of characteristic $0$. We know that a diagonal automorphism $\phi_h$ of $G$ is of the form $g\mapsto hgh^{-1}$, where $h\in \hat H$ and $\hat H$ normalizes $G$. Note that $\hat H:=\{h(\chi)\mid \chi: \mathbb Z \Phi \rightarrow k^* \}$, where $\Phi$ is the corresponding root system and $\mathbb Z \Phi$ is the root lattice.

Let $G'$ be the twisted Chevalley group over the field $k$. I would like to determine diagonal automorphisms for $G'$. So for this we have to determine the analogues group $\hat H'$ which normalizes $G'$.

I am following Robert Steinberg's Yale's lecture notes on "Lectures on Chevalley Groups". At page $106$ (before Theorem $36$), he wrote that for the twisted groups diagonal automorphisms can be defined analogous to the untwisted case.

I am unable to determine what is the group $\hat H'$.

Thank you for your kind help.