= Solution
The https://www-thphys.physics.ox.ac.uk/people/AndreiStarinets/TUTORIALS_NOTES_GR/gr-cosmology-exam-paper-2011-solutions.pdf#page=27[official notes, PDF p. 27, handwritten p. 20] derive the early-time <photon> count, its vanishing limit, and the failure of <thermal equilibrium> when the count is of order one. Two details need supplementing.
First, their constant $N_{\gamma0}$ is the normalization obtained by extrapolating the early-time <particle horizon> formula to $z=0$. It is not the count within today's actual <particle horizon>. Let $N_0=N_\gamma(t_0)$ and let $r_{h0}$ be that horizon's physical radius. Using the paper's early-time approximation gives
$$
\boxed{1+z_*\simeq N_0^{1/3}\frac{c}{H_0r_{h0}}\sqrt{1+z_{\rm eq}},\qquad N_\gamma(z_*)=1.}
$$
For a present matter-dominated approximation, $r_{h0}\simeq2c/H_0$. Alternatively, retaining both components in the flat model exactly, $\Omega_R=1/(2+z_{\rm eq})$, and integrating the <particle horizon> gives
$$
r_{h0}=\frac{2c}{H_0}\frac{\sqrt{2+z_{\rm eq}}}{\sqrt{2+z_{\rm eq}}+1},\qquad
\boxed{1+z_*\simeq\frac{N_0^{1/3}}2\left(\sqrt{2+z_{\rm eq}}+1\right).}
$$
The second boxed formula uses the exact two-component normalization and the radiation-dominated early-time limit. Its large-$z_{\rm eq}$ leading term agrees with the first expression and $r_{h0}\simeq2c/H_0$.
Second, a <particle horizon> containing only order-one <photons> cannot support a smooth, locally representative radiation density: discrete-particle fluctuations are of the same order as the mean, and a single photon's direction is strongly anisotropic. Thus the fluid description assuming <spatial homogeneity> and <cosmological isotropy> is not self-consistent on that causal scale. This does not by itself disprove an ensemble-averaged cosmological symmetry; it limits the extrapolation of the smooth <thermal equilibrium> description.
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