 Solution

A complete pairwise tensor contraction of copies of $A$ is represented by closed index cycles. Each cycle is a trace $\mathrm{tr}(A^r)$, possibly with a sign from reversing an index pair. If a contraction is internal to one copy, it contains the already vanishing $\mathrm{tr}A$. Otherwise an odd total number of copies forces at least one cycle of odd length, since a sum of even lengths is even. That trace vanishes by the transposition argument in part (ii). Therefore every such complete contraction is zero. This proves vanishing odd complete contractions of an antisymmetric matrix without assuming that all contractions form a single trace.