Let $m\ge 2$, and let $G={\rm SO}^*(4m)$ denote the "quaternionic" real form of the special orthogonal group ${\rm SO}(4m,\mathbb C)$ of type ${\sf D}_{2m}$.
Let $\tau\in{\rm Aut}_{\Bbb R}(G)$ be a *real* automorphism of $G$, that is,
an automorphism defined over $\Bbb R$.
My Galois-cohomological calculations suggest that then $\tau$ is an *inner* automorphism.

Question.Is it true that, although $G$ does haveouter automorphisms, it has nocomplexouter automorphisms?real

**Clarification:** I regard $G={\rm SO}^*(4m)$ as an algebraic group over $\Bbb R$. By a complex inner automorphism of $G$ I mean an element of the group ${\rm Inn}(G)(\Bbb C)$, where ${\rm Inn}(G)=G/Z(G)$. By a real inner automorphism of $G$ I mean an element of ${\rm Inn}(G)(\Bbb R)$.

"Gundogan, The component group of the automorphism group of a simple Lie algebra and the splitting of the corresponding short exact sequence", J. Lie Theory 20 (2010), no. 4, 709-737. But I'm not sure what the conclusion is. Apparently, the automorphism group of the real Lie algebra, as Lie group, has 2 components. But I don't see if the resulting homomorphism of the component group to the complex Out is an isomorphism or is the trivial map; this might come from a more careful reading. $\endgroup$ – YCor May 11 '18 at 10:21algebrahere, right? Otherwise I do not know how to make sense of a Lie-group homomorphism being defined 'over a field'. But maybe there is a definition I don't know about? $\endgroup$ – Vincent May 11 '18 at 11:07algebraic group$G$ over $\Bbb R$ with the group of real points $G(\Bbb R)={\rm SO}^*(4m)$. However, if you answer for the Lie algebra, I will be quite happy. $\endgroup$ – Mikhail Borovoi May 11 '18 at 11:23