If $A \in \text{End}(\bigwedge^k \mathbb{R}^d)$ equals $\bigwedge^k B$ for some complex matrix $B$, does it have a real source?

Question

Let $1<k<d$ be an integer. Let $A \in \text{End}(\bigwedge^k \mathbb{R}^d)$, and suppose that $A=\bigwedge^k B$ for some complex $B \in \text{End}(\mathbb{C}^d)$.

Does there exist $M \in \text{End}(\mathbb{R}^d)$ such that $A=\bigwedge^k M$?

In other words: Does every real image of the complex exterior map $$ \psi:\text{Hom}(V,V) \to \text{Hom}(\bigwedge^kV,\bigwedge^kV) \, \, \, \, , \, \, \,\psi(A)=\bigwedge^k A $$ has a real source?

Comment: A positive answer to this question would settle this question.

Answer 1

Counterexample: $d=3$, $k=2$, $A = -I$, $B = iI$. There is no real $M$ because $\det \bigl(\bigwedge^2 M \bigr) = \det^2 M$ and $\det(A) = -1$.