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. Also, I am following "Simple groups of Lie type" by Roger Carter.

I am not sure what is the precise group $\hat H'$?

If we define $\hat H'$ as follows: $$\hat H':=\hat H \cap N_{\mathrm{Aut}(\mathcal L_k)}(G'),$$ then by definition $\hat H'$ normalizes $G'$. Is it true that each of the diagonal automorphisms of $G'$ can be seen as a map

$$\psi_{h'}(g')=h'g'h'^{-1},$$ where $h'\in \hat H'$ and $g'\in G'$?

Thank you for your kind help.