$\newcommand{\Cof}{\operatorname{cof}} \newcommand{\id}{\operatorname{Id}}$ Let $V$ be a real oriented $d$-dimensional vector space ($d>2$). Let $2 \le k \le d-1$ be fixed.
Consider the following map: $$\psi:\text{GL}^+(V) \to \text{GL}(\bigwedge^{k}V,\bigwedge^{k}V) \, \,, \, \, \psi(A)=\bigwedge^{k}A,$$
where $\bigwedge^{k} V$ is the $k$-th exterior power of $V$.
Is $\psi$ an immersion?
Note that $\psi$ is injective and smooth. My motivation is connected to this question.
Edit:
Here are some observations: First, using the multiplicative nature of $\psi$, we can reduce everything to the identity. (Since $d\psi_A(B)=\psi(A) \circ d\psi_{\id}(A^{-1}B)$).
Let's consider for a moment the case $k=2$: In that case $$ \big(d\psi_{\id}(B)\big)(e_i \wedge e_j)=e_i \wedge Be_j+Be_i \wedge e_j. \tag{1}$$
So, we have reduced the question into showing that if equation $(1)$ holds for every two vectors $e_i,e_j$, then $B=0$. (Of course, it suffices to take $e_i,e_j$ to be part of a given basis for $V$).
I have also proved that $\text{trace} B=0$ (In general $d\psi_A(B)=0 \Rightarrow \langle \Cof A,B \rangle=0$ and $\Cof A=\id$).
Since the answer is positive for $k=d-1$ (see explanation below), we see that the first non-obvious case is $k=2,d=4$.
A proof the answer is positive for $k=d-1$:
In that case $\bigwedge^{d-1}A$ is essentially the cofactor matrix of $A$, $\Cof A$. Then we have the following formula for its derivative:
$$d(\Cof)_A(B) = (A^{T})^{-1}\big(\langle \Cof A , B\rangle \cdot \id - B^T \circ \Cof A \big),$$
so $d(\Cof)_A(B)=0$ implies $\langle \Cof A , B\rangle \cdot \id = B^T \circ \Cof A$. Taking traces we get $$d \langle \Cof A , B\rangle = \langle \Cof A , B\rangle \Rightarrow \langle \Cof A , B\rangle =0,$$ which in turn implies $$ B^T \circ \Cof A =0 \Rightarrow B=0.$$
Note that $\psi(A \circ B) = \psi(A) \circ \psi(B);$ so $\psi$ is a homomorphism of Lie groups.
Proposition. Any injective Lie group morphism $\psi : G \to H$ is an immersion.
Proof. Since $\exp_H \circ D\psi_\mathrm{id} = \psi \circ \exp_G$ and exponential maps are diffeomorphisms near the identity, injectivity of $\psi$ implies injectivity of $D\psi_\mathrm{id}.$ By homogeneity we conclude that $D\psi$ is injective everywhere.