Is it possible to find a set of $100$ (or $200$) points on the boundary of a cube such that this set remains fixed under all rotations which leave the cube fixed ?
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For $200$, just put $18$ equally spaced points on each side. You get $12\cdot 18-16=200$ points. I subtracted $16$ because the $8$ corners belong to three sides each.
For $100$ I don't think it's possible. All corners are the same (they all have a point or none of them has one). All interiors of sides have the same number of points, and all interiors of faces have the same number of points, meaning that we must either have $6|100$ or $6|100-8=92$, and none of these is true.
Is it Ok?