Understanding addition of angular momentum for six spin-1/2 nuclei in a molecule

In this paper Matthew Fisher proposes the use of so-called Posner molecules for quantum processing in the brain. I am trying to understand the arguments at the beginning of Section F (Section 6 in the published version here):

Consider first the spin and rotational states of a single Posner molecule. Quantum chemistry calculations find a cubic arrangement for the calcium ions (eight at the cube corners and one at the cube center) but the phosphate ions on the six faces reduce the cubic symmetry to $S_6$, with one 3-fold symmetry axis along a cube diagonal (see Fig. 1)[22–24,16]. The quantum states of the six phosphorus spins can be labeled by the total spin $I_{mathrm{tot}} = 0,1,2,3$, and by a "pseudo-spin", $sigma = 0, pm 1$, encoding the transformation properties under a 3-fold rotation, $vert sigma rangle rightarrow omega^sigma vert sigma rangle$ with $omega = e^{i2pi/3}$. The quantum state with both spin and rotations can be expressed as,
$$vert Psi_mathrm{Pos}rangle = sum_sigma c_sigma vert psi_sigmarangle vert sigma rangle tag{3}$$
with the choice of normalization $langle sigma vert sigma rangle = langle psi_sigma vert psi_sigma rangle = 1$ and $sum_sigma vert c_sigma vert^2 = 1$. The orbital wavefunction $psi_sigma(varphi)$ depends on the angle $varphi$ of rotation about the 3-fold symmetry axis. Fermi statistics requires $vert Psi_{mathrm{Pos}} rangle$ be invariant under a 120 degree rotation that interchanges the positions and spins of the phosphorus ions, [20] imply, $psi_sigma(varphi + 2pi/3) = bar{omega}^{,sigma} psi_sigma(varphi)$ with $bar{omega} = omega^*$. The nuclear spin and rotational states are thus quantum entanglement in the Posner molecule.

My understanding of the setup is the following: we have six spin-1/2 fermions arranged in an octahedron (at the face centers of a cube). Due to the presence of other atoms, the octahedral symmetry is reduced such that there is only a single $C_3$ rotational symmetry about a body diagonal (we could imagine "squishing the cube along the body diagonal" slightly so as to break the cubic symmetry). Here are my questions:

(1) Fisher says that the cubic symmetry is reduced to $S_6$. I do not think I am following his argument: $S_6$ is the permutation group of six elements, of order 64 ($2^6$). Cubic symmetry is the octahedral group $O_h$ of order 48. So I am not understanding where $S_6$ is coming from here.

(2) I am having trouble with the total spin $I_{mathrm{tot}}=0,1,2,3$. I understand we are adding six spin-1/2 angular momenta (with a total of $2^6$ states). I am used to adding angular momenta of two particles, but now I’m a bit lost with six. With the allowed values of $I_{mathrm{tot}}$, it seems that there are only $sum_{I=0}^3 (2I+1) = 16$ total states, not 64 which we have for 6 spin-1/2 particles.

The decomposition is not quite as simple as I thought for more than two angular momenta, it is explained here, the proper decomposition is actually (7) $oplus$ 5 (5) $oplus$ 9 (3) $oplus$ 5 (1), where the notation is (multiplicity)(2s+1).

Fundamental question:

[This is the biggest confusion I have and the main thing I would like answered]

(3) I completely fail to grasp in what way $sigma$ is a pseudo-spin, or what role it is playing in this decomposition. I assume this has something to do with irreps of the symmetry group for the $C_3$ axis (The $C_3$ rotation is represented by a unitary operator with eigenvalues $1, exp(pm i2pi/3)$, the symmetry group is $S_3$.). It seems to me that $sigma$ should be tied to the orbital angular momenta, since it is related to the eigenvalues of a rotation operator, yet the implication I get is that it is somehow related to the spin.