Dotting a Sphere What is the maximum number of points that you can place symmetrically on the surface of a sphere which cannot all be contained within two planes? "Symmetrically" here means that, for any given start point and any given target point, you can rotate the sphere around to get the same set of point positions (over all points, not just start and target) with start now at target. My current maximum uses the bucky-ball with 60 points. (I require more than two planes to avoid 2D circle solutions, with mirroring. I still might be missing a loophole.)

I'm not sure what your question is exactly, but I interpret it as this. Consider a finite set $A$ of points on the sphere $S^2$. We want it to have the property that for each two points $a$, $b\in A$, there is a rotation $R\in O(3)$ with $Ra=b$ and $R(A)=A$.

So each of these $R$ lies in the proper isometry group $G$ of $A$. This is a finite group (unless $|A|\le2$), and it acts transitively on $A$. So $|A|\le|G|$. The classification of finite subgroups of $O(3)$ is known. They are cyclic, dihedral, tetrahedral, octahedral and icosahedral.

I am sure the cyclic and dihedral groups will give you the configurations you want to avoid, with points on two lines.

The largest polyhedral group is icosahedral of order $60$. I suspect $60$ is your maximum number of points.