= Solution
For each rotation, <unitarity> gives
$$
\|U_{ij,\theta}F-F\|_2^2
=2\|F\|_2^2-2\operatorname{Re}\langle F,U_{ij,\theta}F\rangle.
$$
Averaging and using self-adjointness of $Q$ yields the <Dirichlet form of the Kac collision operator>
$$
\boxed{\langle F,(I-Q)F\rangle
=\frac1{4\pi C_N}\sum_{i<j}\int_0^{2\pi}\int_{\mathbb R^N}
|F(R_{ij,\theta}\mathbf v)-F(\mathbf v)|^2\,d\mathbf v\,d\theta.}
$$
The velocity <integral> is necessary: its omission from the printed right-hand side would leave a function of $\mathbf v$ rather than a scalar. This identity applies to every $L^2$ function, with complex modulus when necessary, and is nonnegative.
If $(I-Q)F=0$, every nonnegative angular <integral> is zero, so $\|U_{ij,\theta}F-F\|_2=0$ for almost every angle. Strong continuity in angle extends equality to every angle. The coordinate-plane <Givens rotations> generate the <special orthogonal group> $SO(N)$, hence $F$ is invariant in $L^2$ under every element of this group. To identify its shape rigorously despite almost-everywhere representatives, average over the normalized <Haar measure> of $SO(N)$. This averaging leaves $F$ unchanged, while transitivity of the rotation group on each sphere makes the average a <radial function>. Thus $F(\mathbf v)=\phi(|\mathbf v|)$ almost everywhere.
Conversely, every <radial function> is fixed by every coordinate-plane rotation, and so by $Q$. Therefore
$$
\boxed{\ker(I-Q)=\{F\in L^2(\mathbb R^N):F(\mathbf v)=\phi(|\mathbf v|)\ \text{a.e.}\}.}
$$
This is the <radial kernel of the Kac collision operator>. The rotation correction is essential to this conclusion: with the literal printed map, even $F(\mathbf v)=e^{-|\mathbf v|^2}$ in dimension two is not fixed. At $\mathbf v=(0,1)$ its printed-map angular average is the average of $e^{-\sin^2\theta}$, strictly greater than its value $e^{-1}$.
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