= Solution
Write $F=F_N$ and integrate the <Kac master equation> over $v_2,\ldots,v_N$. For a pair $i,j\geq2$, the rotation acts only on integrated variables. Its unit <Jacobian determinant> makes the integrated gain identical to the integrated loss, so all those pairs cancel.
The only remaining pairs are $(1,j)$, $j=2,\ldots,N$. For such a pair, first integrate over every variable except $v_1$ and $v_j$. This yields the corresponding two-coordinate <marginal distribution> evaluated at the rotated pair. Permutation symmetry of $F_N$ makes all $N-1$ resulting <integrals> identical to the one for $(1,2)$. The coefficient is
$$
\frac{N(N-1)}{2\pi\binom N2}=\frac1\pi.
$$
Consequently the <Kac marginal evolution equation> is
$$
\boxed{\partial_t\Pi_1(F_N)(v_1)
=\frac1\pi\int_{\mathbb R}\int_0^{2\pi}
\left[\Pi_2(F_N)(v_1\cos\theta+v_2\sin\theta,
-v_1\sin\theta+v_2\cos\theta)-\Pi_2(F_N)(v_1,v_2)\right]d\theta\,dv_2.}
$$
The time argument has been suppressed on the right. The loss is consistent with normalization, since $\int\Pi_2(v_1,v_2)dv_2=\Pi_1(v_1)$. Under the printed definition $k<N$, this use of $\Pi_2$ requires $N\geq3$. For $N=2$ the same formula holds with the natural extension $\Pi_N(F_N)=F_N$.
This identity is exact and generally unclosed. Replacing the two-coordinate <marginal distribution> by the product of one-coordinate marginals would produce the quadratic collision equation associated with <Kac chaos>. Permutation symmetry alone does not imply that product approximation.
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