Let $G$ be a group, $(\rho,V)$ a finite dimensional real representation of $G$, and $W$ a subspace of $V$ of dimension $k$.
Assume that $\Lambda^k W$ is a $G$-invariant (and one dimensional) subspace of $\Lambda^k V$.
Does it mean that $W$ is $G$ invariant?
Note: $\Lambda^k$ stands for the $k$-th exterior power.
The answer is yes : assume that $g(w)\not\in W$ for some $w\in W$ and some $g\in G$. Take a decomposition $V = W\oplus W'$ (as vector spaces of course, not as $G$-modules) such that $g(w)\in W'$. Take a basis $(e_i)$ of $V$ that is adapted to this decomposition. You may assume that $e_1 = w$ and $g(e_1) = e_{k+1}$.
Then $e_{k+1}\wedge g(e_1\wedge \dots\wedge e_k) = e_{k+1}\wedge e_{k+1}\wedge (\dots) = 0$. But if we had $g(\Lambda^k W)\subset \Lambda^k W$ then $x:= g(e_1\wedge \dots\wedge e_k)$ would be a non-zero element of $\Lambda^k W$ so $e_{k+1}\wedge x$ would be non-zero in $\Lambda^{k+1}V$.