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.
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]