A curious identity involving a covariant of binary cubic forms

As Abdelmalek notes, if $G$ is to be the cubic covariant $G_F$ of another form $F$, and has a nonzero discriminant of the expected shape $\Delta(G)=3^6n^3$, then $F$ must divide $G'=G_G$ - indeed the only candidate (comparing discriminants and exploiting homogeneity) is $F=-G_G/(3^6n^2)$, which will have rational coefficients when $n$ is rational, and integral coefficients when (and only when) those of $G_G$ are integral multiples of $3^6n^2$.

And when this is the case, this $F$ does indeed work: $$G_F = -G_{G_G}/(3^6n^2)^3 = 3^6\Delta(G)^2G/(3^{18}n^6) = 3^6(3^6n^3)^2G/(3^{18}n^6) = G\,.$$