I wonder whether anyone knows a reference for the following result? I can give a shortish proof, but would prefer to cite the literature if possible.
Theorem Let $V=F^n$ be an $n$-dimensional $F$-space where $F$ is a field. The action of ${\rm GL}(V)$ on $\Lambda^2(V)$ induces a homomorphism $\phi_n\colon{\rm GL}_n(F)\to{\rm GL}_d(F)$ of matrix groups where $d=\binom{n}{2}$. The kernel of $\phi_n$ is ${\rm GL}_1(F)$ if $n=1$, ${\rm SL}_2(F)$ if $n=2$, and $\langle -I_n\rangle$ if $n\ge3$.
I will give a proof if there is no easy reference and one is requested.
I can't think of a reference for the exact result, but two possible short arguments. Maybe something like this is what you had in mind, so this might not answer your question. But in any case, this is too long for a comment.
Proof 1: there are not many normal subgroups of $GL_n(F)$ in general, so you can just check from this what the kernel has to be.
Proof 2: in $\wedge^2(V)$, for nonzero elements $v \wedge w = v' \wedge w'$ if and only if $\langle v,w \rangle = \langle v', w' \rangle$ in $V$. This is basically the fact that the Plücker map is injective, which has textbook references.
For $n = 2$ the map $\phi_n$ is the determinant map, so the result is clear.
Assume $n > 2$. If $g \in \operatorname{ker} \phi_n$, then $g$ leaves every $2$-dimensional subspace of $V$ invariant. This implies that every line $\langle v \rangle$ is invariant under $g$. Indeed, let $\{v,v',v''\}$ be a linearly independent set. Since $\langle v,v' \rangle$ and $\langle v,v'' \rangle$ are $g$-invariant, so is their intersection $\langle v \rangle$.
A linear map which leaves every line invariant must be a scalar, from which the result follows.