There might be bilateral symmetry; I haven't actually looked closely enough to be completely sure. (One thing I haven't yet arranged is a viewing utility that would allow me to rotate one of these solids at whim so I could look at it from all angles. The maths is easy enough, but the UI work sounds too much like effort.)
I've been vaguely wondering about taking one of these point sets (no need to use the output polyhedron) and attempting to find all its symmetries algorithmically. There aren't that many possibilities: simply attempt to map each point on to each other point, and then find other points the same distance from those two candidate points and rotate to try to bring another one into alignment, then see whether you've mapped every point on to (or close to) another. It would in principle take O(N^3) time, but in practice more like O(N^2) times a small constant since there wouldn't be many possibilities for the third candidate point; and you'd end up outputting an exhaustive list of permutations of the points, which would of course form a closed group.
If I could figure out a way to take that set of permutations and compute some sort of canonical characterisation of the group they describe (so that you could tell two point sets had the same symmetry group if and only if the output of this program was identical for them), I would try this, and see just how much symmetry there is in these solids. I noticed in particular that the 13-point set does appear to have two axes of reflective symmetry, so I broadly agree with your intuition that some symmetry is to be expected.
no subject
I've been vaguely wondering about taking one of these point sets (no need to use the output polyhedron) and attempting to find all its symmetries algorithmically. There aren't that many possibilities: simply attempt to map each point on to each other point, and then find other points the same distance from those two candidate points and rotate to try to bring another one into alignment, then see whether you've mapped every point on to (or close to) another. It would in principle take O(N^3) time, but in practice more like O(N^2) times a small constant since there wouldn't be many possibilities for the third candidate point; and you'd end up outputting an exhaustive list of permutations of the points, which would of course form a closed group.
If I could figure out a way to take that set of permutations and compute some sort of canonical characterisation of the group they describe (so that you could tell two point sets had the same symmetry group if and only if the output of this program was identical for them), I would try this, and see just how much symmetry there is in these solids. I noticed in particular that the 13-point set does appear to have two axes of reflective symmetry, so I broadly agree with your intuition that some symmetry is to be expected.