Finite-field points of the semistable quiver stack give the Hall groupoid

Under the inherited assumption [A1], the semistable locus is the open subscheme from [F1]. An object of its quotient groupoid over ${\mathbb{F}}_q$ is a tuple of ${\mathrm{GL}}_{d_i}$-torsors together with an equivariant semistable representation. A ${\mathrm{GL}}_n$-torsor is the frame torsor of its associated $n$-dimensional vector space. Every finite-dimensional vector space over ${\mathbb{F}}_q$ has a basis, so that associated space is isomorphic to ${\mathbb{F}}_q^n$ and its frame torsor is trivial. Applying this at the finitely many vertices trivializes the product $G_{\mathbf{d}}$-torsor. Thus every quotient-stack object is isomorphic to a representation on the standard spaces $k^{d_i}$.

Under a chosen trivialization, an equivariant map to ${\mathrm{Rep}}_{\mathbf{d}}^{ss}$ is an arrow-matrix tuple. Two trivialized objects are isomorphic exactly when their tuples differ by change of bases in $G_{\mathbf{d}}({\mathbb{F}}_q)$. If $\mathbf{d}\mathrm{\ne }0$, [F2] identifies the rational tuples in the geometric semistable locus with the finite-field semistable representations. If $\mathbf{d}=0$, the geometric condition is vacuous, the representation space is a point, and $G_{\mathbf{0}}$ is trivial. Its quotient groupoid therefore has one object with trivial automorphism group, matching the zero representation included by the statement's convention. This proves the claimed equivalence in every dimension.

There are finitely many arrow-matrix tuples on the fixed finite vector spaces, and each automorphism group is a subgroup of the finite group $G_{\mathbf{d}}({\mathbb{F}}_q)$. Hence the groupoid is finite. Under the equivalence of step 2.1, the stabilizer of a tuple is exactly ${\mathrm{Aut}}_{{\mathcal{A}}_Q}(M)$. Substituting these stabilizers into the groupoid-cardinality definition in Finite-field stability data for an acyclic quiver gives the displayed sum.