This is a follow-up of this question.
Let $V,W$ be $d$-dimensional real vector spaces, and let $A,B \in \text{Hom}(V,W)$ be non-invertible maps, $1 \le k \le d-1$.
Consider the induced maps $\bigwedge^{k}A,\bigwedge^{k}B :\Lambda_k(V) \to \Lambda_k(W)$. I want to characterize all the pairs $(A,B)$ which satisfy $\bigwedge^k A=\bigwedge^k B \neq 0.$
Partial results:
$(1)$ I prove below that a necessary condition is $\ker A=\ker B, \text{Image} \, A=\text{Image} \, B$. In particular $\text{rank} \,A=\text{rank} \,B$.
$(2)$ Moreover, contrary to the case where $A,B$ are invertible when the only possible pairs are $(A,A),(A,-A)$, in the singular case there is much more freedom:
Take a $3$-dimensional space with a basis $v_1,v_2,v_3$, and define $A,B$ via: $Av_3=Bv_3=0,Av_1=2v_1,Av_2=\frac{1}{2}v_2,Bv_1=\frac{1}{2}v_1,Bv_2=2v_2$.
This is can be generalized into the following sufficient condition:
$A,B$ have identical eigenspaces, null spaces, and equal products of all the nonzero eigenvalues.
Essentially, since $\ker A=\ker B=W$, $\text{Image} \, A=\text{Image} \, B=\tilde W$, the question really seems about the invertible quotient operators:
$\tilde A,\tilde B:V/W \to \tilde W$.
(I think my example above is related to the special case where $k=\text{rank} \,B=\dim (\tilde W)$. The right condition thenseems to be the "determinant" of the quotient operator is $1$).
Proof that $\ker A=\ker B$:
Let $v \in \ker A$ and assume by contradiction that $Bv \neq 0$. Fix some inner product on $V$, in such a way that $v \in (\ker B)^{\perp}$. Since $\dim(\ker B)^{\perp} = \text{rank} {B} \ge k$, there exist $v_1,\dots,v_{k-1} \in (\ker B)^{\perp}$ such that $v,v_1,\dots,v_{k-1}$ are linearly independent.
Then $ 0=\bigwedge^k A(v \wedge v_1 \wedge \dots \wedge v_{k-1})= Bv \wedge Bv_1 \wedge \dots \wedge Bv_{k-1}$, hence $Bv,Bv_1,\dots,Bv_{k-1}$ are linearly dependent, in contradiction to the fact that $B|_{(\ker B)^{\perp}}$ is injective.
This shows $\ker A \subseteq \ker B$. The other direction follows from symmetry.
The proof that $\text{Image} \, A=\text{Image} \, B$ is easy.
*(I am sure there is a cleaner argument with no insertion of an inner product, e.g via quotients).
It seems to me that all the ingredients of a full answer are included in your question. I'll just put them in the right order.
Yes, $A$ and $B$ have the same kernel $U$ and the same image $\tilde{W}$, so we may think of the two invertible operators $$A',B':V/U\to\tilde{W}.$$
Now, you just need to distinguish between two cases. If $\dim\tilde{W}=k$, then a necessary and sufficient condition is that both operators have the same determinant (this is not even a condition, just another way to formulate the equality you are talking about).
If $\dim\tilde{W}>k,$ the answer to your previous question applies again. That is, $A'$ and $B'$ have to differ by a constant $\alpha$ satisfying $\alpha^k=1$.
It is interesting that the first case allows a big space of solutions (something like $\text{SL}(V/U))$, whereas the solutions in the second case are all trivial.