Count the incongruent n-point sets maximising unit distances: does the number tend to infinity? (Erdős #668)

Is it true that the number of incongruent sets of $n$ points in $\mathbb{R}^2$ which maximise the number of unit distances tends to infinity as $n\to\infty$? Is it always $>1$ for $n>3$? Here two point sets are congruent if an isometry of the plane maps one onto the other, and a set 'maximises the number of unit distances' if no other set of $n$ points in the plane has more pairs of points at Euclidean distance exactly $1$. (The second question is already answered negatively at $n=4$ — see the background — so the substance is the counting question and its asymptotics.)

Acceptance. FULLY RESOLVES: a proof that the number of incongruent $n$-point maximisers of unit distances tends to infinity as $n\to\infty$, or a proof that it does not (e.g. a proof that it equals $1$ for infinitely many $n$ with an explicit characterisation) — machine-checkable preferred, else a complete written proof. ADVANCES: rigorous determination, for concrete values of $n$, of the exact number of incongruent maximising configurations — this requires (a) a proof or certified exhaustive computation of the maximal unit-distance count for that $n$, (b) enumeration of all extremal configurations up to congruence (not merely graph isomorphism), and (c) reproducible code plus realizability certificates; extending such determinations beyond the range described in the background, or upgrading the isomorphism-level counts described there to congruence-level counts within it; or an explicit infinite family of $n$ each admitting at least two incongruent maximisers, with proof. Deliver the enumeration code with its certificates and resulting counts (e.g. as an extension or correction of OEIS A385657), or the proof file.