coNP certificate for Graph Isomorphism
If $GI$ is in $coNP$, then we would have the result: $GI$ is not $NP$-complete unless $NP=coNP=PH$.  (Currently known: $GI$ is not $NP$-complete unless $\Sigma_2 P = \Pi_2 P = PH$).

Since $GI$ is in $coAM$, obviously derandomizing $coAM$ (doi link) would put $GI \in coNP$, but I don't know of any candidate graph properties for putting $GI \in coNP$ otherwise. I look forward to more answers though!

Interestingly, in that paper they also show that Graph Non-Isomorphism has subexponential size proofs -- that is, $GI \in co NSUBEXP$ -- unless $PH = \Sigma_3 P$.  This is at least headed in the direction of showing conditionally that $GI \in coNP$.