Let $G$ be a connected reductive group over a field $k$ with separable closure $k^s$ and $\Gamma=\text{Gal}(k^s/k)$. Pick $T\subset B$ maximal split torus and Borel subgroup for $G_{k^s}$. I am aware of two ways of defining actions of $\Gamma$ on the based root datum $\Psi(G_{k^s},B,T)=(X^*(T),\Phi,\Delta,X_*(T),\Phi^{\vee},\Delta^{\vee})$.
The first is called the $*$-action (see Lemma 12.2.4 of https://math.stanford.edu/~conrad/249BW16Page/handouts/249B_2016.pdf). From what I understand, it is this: $\sigma\in\Gamma$ acts on $\chi\in X^*(T)$ by $\sigma \cdot \chi= \sigma_{\mathbb{G}_m,k^s}\circ \chi \circ \sigma^{-1}_{T}$. But this action will generally not preserve the base $\Delta$ (unless $G$ is quasi-split, i.e. has Borel defined over $k$). One can modified this by letting the Weyl group act on $X^*(T)$ to get back to the base.
The second comes from this map $\Gamma \to \text{Aut}(G_{k^s})\to \text{Aut}(\Psi(G_{k^s},B,T))$ given in section 7.3 of https://sites.duke.edu/jgetz/files/2022/04/Graduate_Text.pdf. The first map is given by sending $\sigma\in\Gamma$ to the pull-back of $\text{Spec } k^s \xrightarrow{\sigma}\text{Spec }k^s \leftarrow G_{k^s}$. For the second map (let me just do this for quasi-split $G$ first), $f\in \text{Aut}(G_{k^s})$ acts on $X^*(T)$ by $\chi \mapsto \chi \circ f|_{T}$.
I suspect these two should be the same action of $\Gamma$ on the based root datum, but I have trouble convine myself of this. For instance, when $G$ is quasi-split, the action of $\Gamma$ on $X^*(T)$ already looks so different?
Any help would be much appreciated!