The problem
Suppose $A,B \in M_{m\times n}(\mathbb R)$.
I want to prove that $\text{null}(A)$ is the same as $\text{null}(B)$ if and only if there exists an invertible matrix $C$ such that $A=CB$.
My solution
From the second half, $A=CB$ where $C$ is invertible implies $C$ is full-rank, therefore nullity of $C$ is the trivial zero. Therefore, $\text{null}(CB) = \text{null}(B) =\text{null}(A) $.
From the first side, if $\text{null}(A)=\text{null}(B)$, assuming we can write $A$ as $CB$ we can say $\text{null}(CB)=\text{null}(B)$. This means that the matrix $C$ as a linear transformation does not nullify anything in $B$. Since both $A$ and $B$ are $m \times n$ then $C$ must be $m \times m$. From the rank-nullity theorem, I can know that since $\text{null}(C)=0$ it is full-rank. So $C$ is invertible.
Am I thinking correctly?
Edit: re-wrote the whole question for comprehensibility.
If you mean $A$ and $B$ have the same null space, then their row space must also coincide, as row spaces and null spaces are orthogonal complements. There is a result that matrices have the same row space iff they are row equivalent, i.e., we can obtain one from another by elementary row operations. Let $C$ be the product of corresponding elementary matrices, we get $A = CB$. For the other direction, you can verify that $Ax = 0$ iff $Bx =0$ with some invertible $C$ such that $A=CB$.
Your permutation matrix is a special case of the product of elementary matrices; specifically, row-interchanging ones. Note that there are also row-multiplication and row-addition elementary matrices.