The answer is no for 100, but yes for 200 points. To see this, choose one initial point on the cubes surface and consider all points this point can be mapped to under an isometry (its orbit under the group of isometries). If our initial point is among the selected 100/200 points, then all of those other orbit points must also be among the selected. If the initial point lies on the interior of one of the cubes 6 faces (that means not on any edge), then clearly the cardinality of its orbit must be a multiple of 6, as any orbit points on its initial face will have to duplicated on the other 5 faces (each cube face can be mapped to any other cube face with an isometry). Since the initial point isn't on an edge, no orbit point is on an edge, so none of those orbit points on two different faces can be identical. If the initial point point lies on an edge but is not a cube vertex, then its orbit cardinality is a multiple of 12, as there are 12 edges (each mappable to each other under isometry) on the cube. Finally, if the initial point is a vertex, then its orbit are all the cubes vertices and thus its cardinality is 8. The whole set of selected points is a discrete union of different orbits. As we saw above, all orbits except one have a cardinality divisible by 6. But 100/200 is not divisible by 6, so both such selected sets must contain the one other orbit (vertices) consisting of 8 points. That leaves 92/192 points of the divisible by 6 orbits. But 92 is not divisble by 6, so there can't be any such selected set of 100 points. OTOH, 192 is divisible by 6, it is even divisble by 48: 192=4*48. If you take a point on the interior of a square, not on any symmetry axis, then there are 7 other points this point can be mapped to under an isometry of the square (4 rotations, plus one reflection), giving 8 points in total. Multiplied with the 6 faces of the cube, the oribit of such a point contains exactly 48 points. Just choose 4 such points (there are infinetely many such points) and their orbits will amount to 192 points. Add the 8 vertices and you have the required 200 points.