= Solution
The <Fredholm Kuranishi reduction> has the following form. Let $F$ be a smooth map between real Hilbert spaces with $F(0)=0$ whose derivative $L=DF(0)$ is a <Fredholm operator>. Split the source as $K\oplus V$, where $K=\ker L$, and the target as $R\oplus C$, where $R=\operatorname{im}L$ and $C$ represents the finite-dimensional cokernel. The restriction $L:V\to R$ is a bounded isomorphism. The <implicit function theorem> applied to $P_RF(k+v)=0$ solves it uniquely near zero as $v=v(k)$, with $v(0)=Dv(0)=0$. Thus
$$
\kappa(k)=P_CF(k+v(k)),\qquad
\kappa:K\longrightarrow C,\qquad
\kappa(0)=D\kappa(0)=0,
$$
is a smooth finite-dimensional obstruction map, and \b[the local zero set of $F$ is the graph over $\kappa^{-1}(0)$]. If a compact group acts preserving the problem, average inner products to choose invariant splittings; uniqueness in the <implicit function theorem> makes the construction equivariant.
Apply this to the <vector-bundle curvature> equation near $A$. Complete connections in $L^2_\ell$ and the <unitary bundle gauge group> in $L^2_{\ell+1}$ with integer $\ell\geq3$, so the required multiplication and gauge operations are smooth. The <Coulomb slice for unitary connections> imposes $d_A^*a=0$. Its local existence follows from the gauge-fixing equation: the derivative in the gauge direction is $d_A^*d_A$, invertible on the orthogonal complement of its parallel kernel by the elliptic estimate. The local slice theorem then says that its remaining identifications are precisely by $\operatorname{Stab}(A)$.
On this slice the equation is
$$
F_{A+a}^+=d_A^+a+(a\wedge a)^+=0.
$$
Its kernel is the harmonic representative space $H_A^1$, and its cokernel is $H_A^2$. The hypothesis $H_A^2=0$ makes the obstruction target zero. Therefore the solutions near $A$ are an equivariant smooth graph over $H_A^1$, and the gauge-orbit neighborhood is a neighborhood of zero in $H_A^1/\operatorname{Stab}(A)$.
It remains to compute this representation, not just its dimension. A <reducible SU2 connection> with circle holonomy preserves $E=L\oplus L^{-1}$. In this splitting,
$$
\operatorname{ad}E\cong\underline{\mathbb R}\oplus(L^2)_{\mathbb R},\qquad
\begin{pmatrix}it&z\\-\overline z&-it\end{pmatrix}
\longleftrightarrow(t,z).
$$
The diagonal part is a trivial real connection. Its degree-one deformation space is the space of ordinary harmonic one-forms. To check this directly, if $d^*a=0$ and $d^+a=0$, then $da$ is an exact <anti-self-dual two-form>. <Stokes theorem> gives
$$
\|da\|_2^2=-\int_Xda\wedge da=-\int_Xd(a\wedge da)=0,
$$
so $da=0$ and $a$ is harmonic. This neutral deformation space vanishes because $b_1(X)=0$. Hence all of $H_A^1$ is in the off-diagonal part, where the complex structure of $L^2$ commutes with both operators; it is a complex vector space.
Using part (a), part (b), and $H_A^2=0$,
$$
\dim_{\mathbb R}H_A^1=\operatorname{ind}D_A+\dim H_A^0
=8e(E)-2,
\qquad
\boxed{d=\dim_{\mathbb C}H_A^1=4e(E)-1.}
$$
Parameterize the stabilizer by $u_\lambda=\operatorname{diag}(\lambda^{-1},\lambda)$, with $\lambda\in S^1$. Under our convention $A^u=u^{-1}Au+u^{-1}du$, its action on an off-diagonal perturbation is
$$
\begin{pmatrix}0&z\\-\overline z&0\end{pmatrix}
\longmapsto
\begin{pmatrix}0&\lambda^2z\\-\overline{\lambda^2z}&0\end{pmatrix}.
$$
Consequently the equivariant graph identifies a neighborhood of $[A]$ with a neighborhood of the origin in
$$
\boxed{\mathbb C^{\,4e(E)-1}/S^1,\qquad \lambda\cdot z=\lambda^2z.}
$$
This proves the exact scalar weight requested, with a stabilizer parametrization consistent with the chosen gauge convention. The ineffective kernel is $\{\pm1\}$. The orbit space is also the cone on $\mathbb{CP}^{d-1}$, because the weight-two circle has the same nonzero orbits as the ordinary scalar circle. This is the <local cone at an unobstructed reducible SU2 instanton>.
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