I was playing around with some equations and noticed the following: Let $A$ be an element of $SU(3)$ with components $A_{ij}$. If $C_{ij}$ is the $(i,j)$ cofactor of $A$ then $C_{ij} = \overline{A_{ij}}$.
Now this is probably not surprising. Since $A \in SU(3)$ we know that its rows and columns form an orthonormal basis for $\mathbb C^3$, so that $$ |A_{i1}|^2 + |A_{i2}|^2 + |A_{i3}|^2 = |A_{1j}|^2 + |A_{2j}|^2 + |A_{3j}|^2 = 1, \qquad \text{ for all } i,j \in \{1,2,3\}.$$ Furthermore, since $\det A = 1$, the cofactor expansion of the determinant tells us that for any $i,j$ \begin{align} 1 &= A_{i1} C_{i1} + A_{i2} C_{i2} + A_{i3}C_{i3} \\ &= A_{1j}C_{1j} + A_{2j}C_{2j} + A_{3j}C_{3j}, \tag{1} \end{align} and this corresponds to the fact that the determinant is invariant under choice of expansion. If $C_{ij} = \overline{A_{ij}}$ then everything works out nicely, but it seems reasonable to expect that there could be other solutions to $(1)$. Perhaps the fact that so many equations need to be simultaneously satisfied enforces a rigidity condition?
In any case, I came across this fact via a rather complicated computation (there are bundles and tori involved), but it seems to me to be sufficiently elementary and fundamental that a much simpler proof should exist, and should generalize to $SU(n)$. Maybe I'm just being foolish, but I cannot seem to see it.
We have $A^\ast = A^{-1} = \frac{\operatorname{adj}(A)}{\det(A)} = \operatorname{adj}(A)$. Taking transpose on both sides, we obtain $\bar{A} = \operatorname{adj}(A)^\top$. Hence $\bar{A}_{ij}=C_{ij}$.