Assume these are general 6j symbols with multiplicity labels and "spins" which are not necessarily selfconjugate (say $SU_3$). From a programming viewpoint, in some use cases it would be far more elegant (especially as if the triad is not "legit", the CG symbol is 0) to write
Clebsch$[l1,l2,l3,l4,j1,j2,j3,j4,j5,j6]=$
Triad$[l1,j1,j2,j3,$
$l2,j1',j5,j6',$
$l3,j2',j4',j6,$
$l4,j3',j4,j5']$
=Triad$[t1,t2,t3,t4]$
where j are spins, l multiplicity labels and ' means conjugation. The triads should be sortable: If $t1=[l1,j3,j2,j1]$ is used instead, the 6j symbol should still be reconstructable uniquely from $t1,t2,t3,t4$. This is surely true for six different spins, but...
Give four triads with them and the spins sorted in any order, but correct conjugation (remember: $t1\neq t1'=[l1,j1',j2',j3']$). (Quick check: Does each spin occur as a pair $j,j'$ in the twelve spin arguments of the triads?) Can you uniquely reconstruct the 6j symbol (modulo its symmetries) defined by them? Or is there a "bad" case where two principially different 6j symbols have the same set of sorted triads?
(There are only a finite number of cases from six equal to six different spins, but the conjugation problem made me shriek back from solving it by brute force. EDIT: I wrote a program, which is not yet complete, but the number of different "spin content" of principially different 6j symbols is around ~1000. Multiply with 5 label sets and 30*24 permutations and you get somewhere between 10000 and 100000 triad sets. This seems manageable for a brute force duplicate check.)
