Clearly we can't have three collinear points on a sphere. Look at the "on a sphere" case and assume there is a configuration where they do not all lie on a single sphere.

Any three points define a plane. The locus of points equidistant from these three is a line perpendicular to the plane (through the circumcentre). There are at most two points on such a line which are unit distance from the original three, $P$ and $Q$ say. These are the centres of two unit circles, and one of the remaining two points must lie on each sphere.

Note that $P$ and $Q$ are related by a reflection in the original plane. There are five sets of four points in the original configuration. Each set defines a unit sphere, and if two spheres are the same, then all are. So there are five spheres and the centres are related by reflections in the planes defined by triangles.

*[From here is a bathtime intuition which, per comments doesn't work. However note that each pair of centres is related by a reflection, which may not map other centres to centres, and therefore creates an infinite group. I thought I could see the group acting on the centres, but I can't make it work - thanks for comments to put me right]*

There is just one group of order 5 - cyclic - and this would imply that the centres of the five spheres formed a regular pentagon. Since this does not provide a suitable configuration, none exists. [Would need three points in each of five planes meeting in a single line, no three collinear]