Suppose that $A,B\in Mat_{m\times n}(\mathbb{C}),m\leq n.\det(A\overline{A^{'}})\ne 0,\det(B\overline{B^{'}})\ne 0.$
Show that $$ \exists C\in Mat_{m\times m}(\mathbb{C}),s.t. B=CA. \Longleftrightarrow $$$$\exists \lambda\in \mathbb{C},s.t. \forall 1\leq i_{1}< i_{2}<\cdots<i_{m}\leq n,$$ $$ \det (B\begin{pmatrix} 1& 2& \cdots& m\\ i_{1}& i_{2}& \cdots& i_{m} \end{pmatrix})=\lambda\cdot\det( A \begin{pmatrix} 1& 2& \cdots& m\\ i_{1}& i_{2}& \cdots& i_{m} \end{pmatrix}).$$
$A\in Mat_{m\times n}(\mathbb{C})$ means $A$ is an $m$-by-$n$ complex matrix.
$\overline{A^{'}}$ means the conjugate transpose of $A$.
$A\begin{pmatrix} 1& 2& \cdots& m\\ i_{1}& i_{2}& \cdots& i_{m} \end{pmatrix} $ is a submatrix of $A$ ,in which the set of row indices is $\{1,2,\cdots ,m\} $ and the set of column indices is$\{i_{1},i_{2},\cdots ,i_{m}\}$.
$\Rightarrow_{.}$ It is easy to verify.
$\Leftarrow_{.}$ In order to simplify this question, I let $m=2,n=3.$$A=(a_{ij})_{2\times 3},B=(b_{ij})_{2\times 3}.$ Then $$\det\begin{pmatrix} b_{11}& b_{12} \\ b_{21}& b_{22} \end{pmatrix}=\lambda\cdot \det\begin{pmatrix} a_{11}& a_{12} \\ a_{21}& a_{22} \end{pmatrix},$$ $$\det\begin{pmatrix} b_{11}& b_{13} \\ b_{21}& b_{23} \end{pmatrix}=\lambda\cdot \det\begin{pmatrix} a_{11}& a_{13} \\ a_{21}& a_{23} \end{pmatrix},$$ $$\det\begin{pmatrix} b_{12}& b_{13} \\ b_{22}& b_{23} \end{pmatrix}=\lambda\cdot \det\begin{pmatrix} a_{12}& a_{13} \\ a_{22}& a_{23} \end{pmatrix}.$$
How can I find a matrix $\begin{pmatrix} c_{11}&c_{12} \\ c_{21}&c_{22} \end{pmatrix},$ satisfying the following equations: $$ \begin{cases} \qquad\qquad\qquad\quad\lambda=\det\begin{pmatrix} c_{11}&c_{12} \\ c_{21}&c_{22} \end{pmatrix}; & \\ \begin{pmatrix} b_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23} \end{pmatrix}=\begin{pmatrix} c_{11}&c_{12} \\ c_{21}&c_{22} \end{pmatrix}\begin{pmatrix} a_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23} \end{pmatrix}.& \end{cases} ?$$
Let $A = [A_1\;A_2]$ and $B = [B_1\;B_2]$ and assume that $A_1$ and $B_1$ are invertible. If this is not the case, rearrange the columns. Set $D := \det(A_1)$. Then $\det(B_1) = \lambda D$. Define $C := B_1A_1^{-1}$. Then $CA_1 = B_1$.
Now, let $a$ be an arbitrary column in $A_2$ and $b$ the corresponding one in $B_2$. We have to prove that $Ca = b$. Let $x := A_1^{-1}a$ and $y := B_1^{-1}b$. Let $j\in\{1,\ldots,m\}$ and let $A_1'$ be the matrix $A_1$ with missing $j$-th column. Define $B_1'$ accordingly. Then (denoting by $A_1^k$ the $k$-th column of $A_1$) \begin{align*} \det(A_1'|a) &= \det(A_1'|A_1x) = \det(A_1'|\sum_{k=1}^mx_kA_1^k) = \sum_{k=1}^mx_k\det(A_1'|A_1^k)\\ &= x_j\det(A_1'|A_1^j) = (-1)^{m-j} x_j\det(A_1) = (-1)^{m-j}Dx_j. \end{align*} Analogously, $$ \det(B_1'|b) = (-1)^{m-j}y_j\det(B_1) = (-1)^{m-j}\lambda Dy_j. $$ Therefore, $$ (-1)^{m-j}\lambda Dy_j = \det(B_1'|b) = \lambda\det(A_1'|a) = (-1)^{m-j}\lambda Dx_j. $$ This implies $y = x$ and thus $Ca = CA_1x = B_1 x = B_1y = b$.