Does the action of a linear map on $k$-dimensional subspaces determine it up to scaling?

The equality of kernels can be shown to always hold and then the argument via isomorphisms on quotient spaces works through:

Pick a basis $v_1,\ldots, v_n$ of $\ker A\cap \ker B$. Pick $v^A_1,\ldots,v^A_m$ such that together with the $v_i$, they form a basis of $\ker A$. Pick $v^B_1,\ldots,v^B_m$ such that together with the $v_i$, they form a basis of $\ker B$. (That the same $m$ occurs as for the $v^A_i$ follows from lemma 2). These $n+2m$ vectors are linearly independent: If $\sum_i c_iv_i+\sum_ic^A_iv^A_i+\sum_ic^B_iv^B_i=0$, then apply $A$ to find $\sum_ic^B_iv^B_i\in \ker A\cap \ker B$, hence all $c_i^B=0$. Likewise all $c_i^A=0$ and then all $c_i=0$. Hence we can pick $u_1,\ldots, u_{d-n-2m}$ such that $$\tag1v_1,\ldots, v_n, v^A_1,\ldots,v^A_m, v^B_1,\ldots,v^B_m, u_1,\ldots, u_{d-n-2m}$$ form a basis of $V$. Note that the $Av_i^B$ and $Au_i$ form a basis of $V/\ker A$, hence are linearly independent. Similarly, the $Bv_i^A$ and $Bu_i$ are linearly independent. Let $W$ be the subspace spanned by $k$ of the vectors in $(1)$. Then $\dim A(W)$ is the number of vectors picked from $v_1^B\ldots v_m^B, u_1,\ldots, u_{d-n-2m}$ and $\dim B(W)$ is the number of vectors picked from $v_1^A\ldots v_m^A, u_1,\ldots, u_{d-n-2m}$. By the given property, these numbers must be equal and hence also the same number of vectors were picked from the $v_i^A$ as from the $v_i^B$. If $m>0$ and $k<d$, it is clearly possible to violate this condition. We conclude $m=0$, i.e,

$$ \ker A=\ker B.$$

Let $D=\ker A=\ker B$ and $H=A(V)=B(V)$. If $D$ has codimension $\le 1$, then $\dim \operatorname{Hom}(V/D,H)\le 1$ and so $A,B$ are linearly dependent. In all other cases, let $k'=\max\{1,k-\dim D\}$ and consider the isomorphisms $\tilde A,\tilde B\colon V/D\to\operatorname{im}(A)$ that $A,B$ induce. For any $k'$-dimensional subspace $\tilde W$ of $V/D$, we find a $k$-dimensional subspace $W$ of $V$ in the $(k'+\dim D)$-dimensional preimage and conclude $\tilde A(\tilde W)=A(W)=B(W)\tilde B(\tilde W)$. As $1\le k'<\dim (V/D)$, lemma 1 shows that $\tilde A$ and $\tilde B$ are linearly dependent, hence so are $A$ and $B$.