With two equilateral triangles, each triangle obviously rotates onto itself. So two of the 6! triples of points satisfy the constraint. How about triples composed of two corners of one triangle and one corner of the other?

Thanks for your question. The triple $j_{1}, j_{2}, j_{3}$ might differ from $l_{1}, l_{2}, l_{3}$ only in order. The ordered triples are distinct. I've updated the question to make this point clearer.

Thanks for the comment. The hexagon corresponding to the six points you described has two-fold rotational symmetry about an axis passing through $r_{1}$. I would like to know whether satisfying the constraint always implies some sort of rotational (perhaps also reflection) symmetry of the points.

I'm only interested in 3D and $n > 3$. A rotationally-symmetric polyhedron with vertices on the unit sphere obviously satisfies the constraint, but I would like to know what can be said in the opposite direction.

That's right. For every $j_{1}, j_{2}, j_{3}$ there is some $l_{1}, l_{2}, l_{3}$ such that the equation is satisfied for some rotation.