Matrices with same columns Consider two linear transformations $L_1, L_2: V \to W$.
Fix a basis of $V$, $W$, and consider $M_1$, $M_2$, the matrices of the aforementioned transformations w.r.t said basis.
Suppose you can obtain $M_2$ from swapping columns in $M_1$.
How are $L_1$ and $L_2$ related? (Besides having the same image)
 A: If $M_2$ is obtained from $M_1$ by swapping the first two columns (for example), and if we denote by $(v_1,\dots,v_n)$ the basis of $V$, then we have 
$$L_2(v_1)=L_1(v_2)\text{ and }L_2(v_2)=L_1(v_1).$$
This is because in the matrix of a linear map $L:V\to W$, the $j$th column contains the coordinates (in the basis of $W$ chosen) of $L(v_j)$, the image of the $j$th basis vector of $V$. Hence, the first equality I wrote above means that the first column of $M_2$ is the second column of $M_1$ and the second equality I wrote above means that the second column of $M_2$ is the first column of $M_1$.
Edit: If $\varphi$ is the permutation of the $v_i$'s (e.g. $\varphi(v_1)=v_2$, $\varphi(v_2)=v_1$ and $\varphi(v_i) = v_i\;\forall i\geq 3$ in the example above), then $$L_2 = L_1\circ \varphi.$$ Indeed, one can check that $\forall i \in\{1,\dots,n\},\;L_2(v_i) = L_1(\varphi(v_i))$. 
In fact, since the transpositions generate the symmetric group, this example is sufficient.
A: because the transformation matrix is determined by basis of V and W, if you swap the two column (e.g. i and j) of the matrix, you are swapping two basis $w_i$ and $w_j$ of W. L1 and L2 are the same but different matrix interpretation.
