Let $A$ be an Azumaya algebra over the commutative ring $R$ and let
$\newcommand{\tr}{\operatorname{tr}} \tr\colon A \to R$ be the reduced trace as defined (for example) in Section IV.2 of M.-A. Knus, M. Ojanguren, *Théorie de la Descente et Algèbres d'Azumaya* (doi:10.1007/BFb0057799). Let $g$ be an automorphism of $A$ as ring, but not necessarily as $R$-algebra. Since we may identify $R$ with $Z(A)$, $g$ induces an automorphism of $R$. I need a reference/proof that the following holds:

$\tr(a^g) = \tr(a)^g $ for all $a\in A$.

In trying to prove this myself, I came up with the following.

There is a unique element $t=\sum x_i \otimes y_i \in A\otimes A $ such that $\tr(a) = \sum x_iay_i $ for all $a\in A$. This element has the properties $t^2 = 1$ and $t (a\otimes b) = (b\otimes a) t$. The element $t^g$ has the same property and so $t^g = rt $ for some $r\in R$ and $r^2=1$. This means that $\tr(a^g)= r\tr(a)^g$. It remains to show $r=1$.

Another attempt: It is not difficult to reduce to the case where $A$ has constant rank. Let $S$ be a faithfully flat extension such that $A\otimes S \cong \mathbf{M}_n(S)$. If $g_{|R}$ extends from $R$ to a ring automorphism of $S$, then we can extend $g$ to $A\otimes S$ in an obvious way. Since trace and scalar extensions commute, this reduces the question to the case where $A$ is a matrix ring. In that case we can compose $g$ with the automorphism of $A\cong \mathbf{M}_n(R)$ that acts as $g^{-1}$ on the entries of a matrix. We get an $R$-algebra automorphism of $A$, and for these the result follows from Lemme IV.2.2 in Knus-Ojanguren.

However, I do not know whether there always is a faithfully flat extension $S$ that splits $A$ and such that $g_{|R}$ extends to $S$. (In my application, $g$ has finite order, if that helps.)