The Brauer-Picard 2-category of $SuperVect_{\mathbb R}$ (let's call it $sBrPic_\mathbb R$) is the homotopy fixed points of the Brauer-Picard 2-category of $SuperVect_{\mathbb C}$ w.r.t. the involution given by complex conjugation (let's call that involution $C$).

It is plausible that the Brauer-Picard 2-category of $SuperVect_{\mathbb H}$ (call it $sBrPic_\mathbb H$) is the homotopy fixed points of $sBrPic_\mathbb C$ w.r.t the involution $C\circ J$, where $J$ is the involution defined here: What are the "correct" conventions for defining Clifford algebras?

As explained in the above link, $J$ acts homotopy trivially on $SuperVect_{\mathbb C}$.
The action of $C\circ J$ on the homotopy groups of $sBrPic_\mathbb C$ (which are $\mathbb Z/2,\mathbb Z/2,\mathbb C^\times$) is therefore the same as that of $C$, and so the homotopy fixed points spectral sequence for $C\circ J$ looks the same as that for $C$:

$$
\begin{matrix}
H^2(\mathbb Z/2,\pi_2(sBrPic_\mathbb C)) \\
H^1(\mathbb Z/2,\pi_1(sBrPic_\mathbb C)) & H^1(\mathbb Z/2,\pi_2(sBrPic_\mathbb C)) \\
H^0(\mathbb Z/2,\pi_0(sBrPic_\mathbb C)) & H^0(\mathbb Z/2,\pi_1(sBrPic_\mathbb C)) & H^0(\mathbb Z/2,\pi_2(sBrPic_\mathbb C))\\
\end{matrix}
$$

$$=\qquad
\begin{matrix}
\mathbb Z/2 \\
\mathbb Z/2 & 0 \\
\mathbb Z/2 & \mathbb Z/2 & \mathbb Z/2\\
\end{matrix}
$$

But the differential could be different:
there is room for a $d_2$ differential going from $(1,0)$ to $(0,2)$.

That differential is present iff
$\pi_0(sBrPic_\mathbb H)$ has order four, and also iff $\pi_1(sBrPic_\mathbb H) = 0$.

And indeed, $\pi_1 = 0$ because the ``odd line'' is not invertible in the category $SuperVect_{\mathbb H}$.

So you were right: it looks like $\pi_0(sBrPic_\mathbb H) = \mathbb Z/4$.

My answer raises the question of what is the homotopy fixed points of $J$ acting on $sBrPic_\mathbb C$, or $sBrPic_\mathbb R$?

I have no reason to believe that this is related to a version of $K$-theory.

Do you know a version of $K$-theory related to the category of (non-super) vector spaces? - probably not.
I just don't think that every linear symmetric monoidal category corresponds to a version of $K$-theory.

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user André Henriques