From the ATLAS, I know that the outer automorphism group of the Chevalley group $D_n(q)$, $q=p^f$ for some prime $p$ and some $n$ even and $n>4$, is a semidirect product of three groups, $(C_d \times C_d) \rtimes (C_f \times C_g)$, where $d=(2,q-1)$ (the "diagonal" automorphisms), $f$ is such that $q=p^f$ (the "field" automorphisms) and $g=2$ (the graph automorphisms), so
$$\operatorname{Out}(D_n(q))= (C_2 \times C_2) \rtimes (C_f \times C_2)$$
What I want to know is: when $f=3k$ for some $k \in \mathbb{N}$, does $C_f$ act on $C_2 \times C_2$? Equivalently, do the field automorphisms and the diagonal automorphisms commute?
I am also interested in the $n=4$ case, when
$$\operatorname{Out}(D_4(q))= (C_2 \times C_2) \rtimes (C_f \times S_3)$$
and I ask the same question for $C_f$, but also for $C_3 \leq S_3$.
Best Answer
The subgroup $(C_d \times C_d) \rtimes C_g$ is dihedral of order $8$, and since $C_f$ commutes with $C_g$, it follows that $C_f$ must commute with $C_d \times C_d$, because ${\rm Aut}(C_2 \times C_2) \cong S_3$, not $C_6$.
On the other hand, when $n=4$, the $S_3$ subgroup acts faithfully on $C_d \times C_d$, and the subgroup $(C_d \times C_d) \rtimes S_3$ is isomorphic to $S_4$. This is the only situation in which the outer automorphism group of a finite simple group has derived length $3$.