What is the symmetry of the cuboctahedron (FCC metal)?

The background to this is that I've recently given a tutorial wherein we had to go through the determination of point groups for atoms in various lattices (BCC, FCC/CCP and HCP). BCC and HCP, I have no problem with. However, I've now acquired a copy of the lecturer's "official" solutions and there's a bit of discrepancy with the FCC, that I can't fathom: I get an $O\_\mathrm{h}$ point group, while he gives a $D\_\mathrm{3d}$ point group.


Here goes:


The coordination geometry of an atom in the FCC structure (assuming one atom/lattice point and only one species of atom, as in an FCC metal) is a cuboctahedron:


![coordination geometries](https://i.stack.imgur.com/61CwA.png), ![cuboctahedron](https://upload.wikimedia.org/wikipedia/commons/thumb/a/a6/Cuboctahedron.png/240px-Cuboctahedron.png)


By my reckoning we have $3$ $C\_4$ axes, one through each pair of square faces, and a $C\_3$ through each pair of triangular faces.


Following the classic flowchart:


![point group flowchart](https://2.bp.blogspot.com/-hURHYxAd53w/TaUKze5BCgI/AAAAAAAAACw/tkOVBy4s3rI/s1600/image040.jpg)


The number of high-symmetry rotational axes is surely enough to exclude $D\_\mathrm{3d}$ already?


The only thing that makes me think that the lecturer hasn't just made an error is this line from the [Wikipedia page for the cuboctahedron](https://en.wikipedia.org/wiki/Cuboctahedron):



> 
> With $O\mathrm{h}$ symmetry, order $48$, it is a rectified cube or rectified octahedron (Norman Johnson).
> 
> 
> With $T\_\mathrm{d}$ symmetry, order $24$, it is a cantellated tetrahedron or rhombitetratetrahedron.
> 
> 
> With $D\_\mathrm{3d}$ symmetry, order $12$, it is a triangular gyrobicupola.
> 
> 
> 


This is no doubt due to deficiencies in my understanding of group theory, but I don't understand how the same shape can simultaneously have multiple point groups, nor how you would select which one is relevant to a particular analysis.


