= Solution
Use the genuine planar <Givens rotation>
$$
\binom{v_i(\theta)}{v_j(\theta)}
=\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}
\binom{v_i}{v_j}.
$$
In the second component the cosine multiplies $v_j$: the repeated $v_i$ in the printed formula is an error. With that printed expression, at $\theta=0$ the pair becomes $(v_i,v_i)$, which does not preserve length or measure. The rotation-based claims require the corrected expression. Also take $N\geq2$, since the normalization by $\binom N2$ is undefined for $N=1$.
Let $C_N=\binom N2$ and $U_{ij,\theta}F=F\circ R_{ij,\theta}$. The <change of variables formula> and determinant one give $\|U_{ij,\theta}F\|_2=\|F\|_2$. Thus each is a <unitary operator>, with adjoint $U_{ij,-\theta}$. The <Kac collision operator> is the average
$$
Q=\frac1{C_N}\sum_{i<j}\frac1{2\pi}\int_0^{2\pi}U_{ij,\theta}\,d\theta.
$$
The <Minkowski integral inequality> gives $\|QF\|_2\leq\|F\|_2$, so $Q$ is bounded. For the <Hilbert space> <inner product>, integration and the angular change $\theta\mapsto-\theta$ give
$$
\langle G,QF\rangle
=\frac1{2\pi C_N}\sum_{i<j}\int_0^{2\pi}\langle U_{ij,-\theta}G,F\rangle d\theta
=\langle QG,F\rangle.
$$
Hence \b[$Q=Q^*$ and $\|Q\|\leq1$]. In fact a nonzero radial <Gaussian function> is fixed by every rotation, showing $\|Q\|=1$. Angular averages can be understood as strong <Bochner integrals>; continuity of rotations in $L^2$ follows first for smooth compactly supported functions, then by density.
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