I haven't worked through the details, but since nobody has answered this fully, I'm going to crib from [this][1] presentation by @JohnBaez to hopefully expand on the comments above.

First, I'm going to state a few facts without proof. The space of bound state solutions to the Kepler problem in three dimensions is the (punctured) cotangent bundle to $S^3$. Thus, the classical question is equivalent to a free particle on $S^3$. I believe this is due to Fock. The $SO(4)$ action arises from the two $SU(2)$ factors in $Spin(4)$ acting on both sides of $S^3 \cong SU(2)$.

With this, we can decompose a la the Peter-Weyl theorem:
$$
L^2(S^3) \cong \bigoplus_i \rho_i \otimes \rho_i^\star
$$
Each $\rho$ is an irrep of $SU(2)$, and the irreps can be labelled by an integer $n$, so the problem decomposes into $n^2$ dimensional irreps of $SO(4)$, which explains the question asked.


  [1]: https://math.ucr.edu/home//baez/hydrogen/4d/hydrogen_4d.pdf