= Solution
Use physical momentum $p_{\mathrm{phys}}$ and conserved-background <comoving momentum> $q=a p_{\mathrm{phys}}$. At fixed $q$, the homogeneous collisionless <phase-space distribution function> is $f_0(q)$, with no spatial or angular dependence and no explicit background time derivative. The printed relation $p_i=aqn_i$ cannot have this meaning: freely propagating physical momentum redshifts as $a^{-1}$, so it should read $p_i=q n_i/a$. Keeping the printed relation literally would give $q=p_{\mathrm{phys}}/a$ and a nonzero zeroth-order redshift, inconsistent with the target equation. The standard definition is positively confirmed by https://arxiv.org/html/astro-ph/9506072[the phase-space convention in Ma and Bertschinger's perturbation treatment].
With this conserved $q$, the orders of the factors are:
* $\partial_t f=\partial_t f_1$ is first order; the zeroth-order collisionless equation has already made $\partial_t f_0=0$.
* $\partial_{x^i}f$ is first order, while $dx^i/dt$ has a zeroth-order free-streaming velocity and first-order corrections. Their product contributes at first order only through that background velocity.
* $\partial_qf=f_0'(q)+\partial_qf_1$ begins at zeroth order; $dq/dt$ begins at first order. Its leading product is $f_0'\dot q_1$.
* $\partial_{n^i}f=\partial_{n^i}f_1$ begins at first order, and $dn^i/dt$ begins at first order because background trajectories keep their direction. Their product is second order and drops out.
Thus the linear <Collisionless Boltzmann equation> is
$$
\boxed{\partial_t f_1+v_0^i\partial_{x^i}f_1+\dot q_1 f_0'(q)=0.}
$$
For photons $v_0^i=n^i/a$. For a particle of mass $m$, $v_0^i=q n^i/[a\sqrt{q^2+a^2m^2}]$. The order counting does not assume that the first-order angular deflection itself vanishes; only its multiplication by an already perturbed angular distribution makes it higher order.
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