There’s a bit of multilinear-algebraic folklore knocking around in my head that I’d like to cite and use but that I can’t track down for the life of me:
Let $n > 2$ be an integer, and let $a,b \in \operatorname{GL}(n,\mathbb{R})$. If $\bigwedge^2 a = \bigwedge^2 b$, then $a=\pm b$.
Note that the condition $\bigwedge^2 a = \bigwedge^2 b$ is equivalent to the equality $C_2(a) = C_2(b)$ of $2$nd compound matrices, but I’ve had no luck so far with what seem to be the usual references for compound matrix lore. So, assuming that this half-remembered lore is correctly remembered, is there a handy citation one can provide for it?
Since $C_2$ is a multiplicative operator, we have $$C_2(A) = C_2(B) \Longrightarrow C_2(A B^{-1}) = C_2(B B^{-1}) = I.$$ So it suffices to show that for an $n \times n$ matrix $M$, $C_2(M) = I$ iff $M = \pm I$. This can be seen as follows: we have $$M_{11} M_{22} - M_{12} M_{21} = 1, M_{11} M_{23} = M_{21}M_{13}, M_{12} M_{23} = M_{22}M_{13}.$$ So if $(M_{13}, M_{23})$ is not the zero vector, then $(M_{11}, M_{21})$ and $(M_{12}, M_{22})$ are both its scalar multiplication, meaning that $M_{11} M_{22} - M_{12} M_{21} = 0$, contradiction! So we have $M_{13} = M_{23} = 0$.
Symmetrically, one can show that all off-diagonal elements are zero. Finally, note that $M_{ii} M_{jj} = 1$ for any $i \neq j$, so it follows that the diagonal elements are either all $1$ or all $-1$.