Automorphism of the Dynkin diagram

Question

I have a question about a passage from Springer's Linear Algebraic Groups (Birkhauser). The setup is: $F$ is an arbitrary field, $G$ is an $F$-split group, and $B$ is a Borel subgroup of $G$ containing a maximal $F$-torus $T$. This corresponds to a choice $R^+$ of positive roots. Let $D$ be the root basis. Let $\mathcal{D}$ be the Dynkin diagram defined by $D$. On page 276, Springer writes, "An automorphism of $G$ stabilizing $B$ and $T$ and fixing the elements of the connected center $C$ induces an automorphism of $\mathcal{D}$."

My question is, if an automorphism of $G$ stabilizes $B$ and $T$, why doesn't it already induce an automorphism of the Dynkin diagram? My reasoning was that if $\sigma$ is the automorphism, then $\sigma$ preserves the minimal $T$-stable subspaces of the unipotent radical $R_u(B)$, namely the root subgroups $U_\alpha$ for $\alpha \in R^+$. This induces a permutation of the positive roots, which stabilizes the set of simple roots. Therefore one gets an automorphism of the Dynkin diagram.

Answer 1

It is true that any automorphism stabilizing $B$ and $T$ induces an automorphism of $\mathcal D$. But Springer continues: "Let $A_0$ be the subgroup of $A$ of automorphisms of $\mathcal D$ obtained in this manner." So your quote is just the introduction for the definition of $A_0$.