Suppose that $k$ is a commutative ring and that $A$ is an Azumaya $k$-algebra. Then there is a well-known morphism from $Aut_{Alg_k}(A)$, the group of algebra automorphisms, to the Picard group $Pic(k)$. One way to phrase it is as follows. Every automorphism $\phi$ determines a $k$-linear autoequivalence $$ \phi^*: Mod_A \to Mod_A $$ However, since $A$ is Azumaya, every $k$-linear autoequivalence of the category of (right) $A$-modules is of the form $M \mapsto M \otimes_k J$ for some $J$ in the Picard group of $k$, unique up to isomorphism. Moreover, this assignment takes composition to tensor product. This assignment is part of the Rosenberg-Zelinsky exact sequence $$ 0 \to k^\times \to A^\times \to Aut_{Alg_k}(A) \to Pic(k). $$ It is possible to express $J$ as $HH^0(A,A^\tau)$ where $A^\tau$ is $A$, but with half of its bimodule structure pulled back along $\phi$.

In particular, given an Azumaya $k$-algebra $Q$, let $A = Q \otimes_k Q$. Then A is also an Azumaya $k$-algebra with automorphism $\phi(a \otimes b) = b \otimes a$, and thus $Q$ determines a (2-torsion) element in $Pic(k)$. This assignment turns out to be Morita invariant and it respects the tensor product; this makes it a (2-torsion) homomorphism from the Brauer group $Br(k)$ to the Picard group $Pic(k)$.

Are there examples of rings $k$ for which this homomorphism is nontrivial?