On a morphism from the Brauer group to the Picard group

It seems to me that the involution of $Q \otimes Q$ that exchanges $a \otimes b$ and $b\otimes a$ is inner, which means that the homomorphism you describe should always be trivial.

Here is the argument, I hope it is correct. Suppose that $Q$ is trivial, that is, there is a projective $k$-module $V$ such that $Q \simeq \operatorname{End}V$. Then $Q \otimes Q \simeq \operatorname{End}(V\otimes V)$. Consider the operator $\tau\colon V\otimes V \simeq V\otimes V$ that exchanges $v \otimes w$ and $w \otimes v$. We can think of $\tau$ as an element of $Q \otimes Q$; conjugation by $\tau$ exchanges $a \otimes b$ and $b\otimes a$.

Now, choose another projective $k$-module $W$ and an isomorphism $Q \simeq \operatorname{End}W$. Consider the induced isomorphism $\phi\colon\operatorname{End}V \simeq \operatorname{End}W$. Then there exists an invertible $k$-module $L$ and an isomorphism $\psi\colon W \simeq L \otimes V$ such that $\psi$ that induces $\phi$. This implies that $\tau$ is independent of the isomorphism $Q \simeq \operatorname{End}V$, and it is a canonical element of $Q\otimes Q$.

If $Q$ is not trivial, choose a faithfully flat extension $k'/k$ such that $Q_{k'}$ is trivial. Because it is canonical, by descent theory the element $\tau' \in Q_{k'}\otimes_{k'}Q_{k'}$ constructed above descends to an element $\tau \in Q \otimes_k Q$ which will induce the involution $Q \otimes Q$ that exchanges $a \otimes b$ and $b\otimes a$.

Here is another way to see that the given map should be zero. It corresponds to a map of sheaves of grouplike $\mathbb{E}_\infty$-spaces $K(\mathbb{G}_m,2)\rightarrow K(\mathbb{G}_m,1)$. All such maps are nullhomotopic (for example by delooping to sheaves of spectra and using a $t$-structure argument).

I would also guess that the map is given by $A\mapsto\mathrm{HH}(A/k)$. Since Hochschild homology is symmetric monoidal, $\mathrm{HH}(A/k)$ is a line bundle. The argument above shows that it is trivial.

On the other hand, I suspect there are derived Azumaya algebras where this is non-trivial. If we look at the derived Brauer sheaf (on the étale site of an ordinary commutative ring) it is an extension $$K(\mathbb{G}_m,2)\rightarrow\mathbf{dBr}\rightarrow K(\mathbb{Z},1).$$ Taking Hochschild homology gives a map of sheaves of grouplike $\mathbb{E}_\infty$-spaces $\mathbf{dBr}\rightarrow\mathbf{dPic}$, where $\mathbf{dPic}$ fits into a fiber sequence $$K(\mathbb{G}_m,1)\rightarrow\mathbf{dPic}\rightarrow K(\mathbb{Z},0).$$ The map $\mathbf{dBr}\rightarrow\mathbf{dPic}$ thus canonically factors through a map $K(\mathbb{Z},1)\rightarrow K(\mathbb{G}_m,1)$ of grouplike $\mathbb{E}_\infty$-spaces. I believe this is the map that sends $1$ to $-1\in\mathbb{G}_m(\mathbb{Z})$. Indeed, the copy of $\mathbb{Z}$ is coming from suspension in the derived category.

To make this concrete, one needs a ring $k$ where the induced map $H^1(k,\mathbb{Z})\rightarrow\mathrm{Pic}(k)$ is non-zero. I don't have an example, but I'd be surprised if this didn't exist.