Let $n,m≥1$, let $\Bbb K$ be a field and let $A,B\in \Bbb K^{m×n}$ be matrices for which the systems of linear equations $Ax= 0$ and $Bx= 0$ have the same solutions.
(A) $C\in \Bbb K^{n×n}$ exists such that $B =A·C$.
(B) There exists $C\in \Bbb K^{m×m}$ such that $B=C·A$.
(C) The kernels of the linear maps determined by $A$ and $B$ are identical.
Why is (A) false and (B) true? And how do I show that their kernels are in fact identical?
We'll start with (C): this is simply a restatement of the assumption. That $A$ and $B$ have the same solutions means that $Ax=0$ for some $x \in \mathbb{K}^n$ if and only if $Bx=0$. But the set of all $x$ such that $Ax=0$ is precisely the definition of the kernel of $A$. So they have the same kernels.
Let's consider statement (B): From the rank-nullity theorem we know that $$ \dim(\ker A) + \dim(\operatorname{im}A) = \dim{\mathbb{K}^n}=n \\ \dim(\ker B) + \dim(\operatorname{im}B) = \dim{\mathbb{K}^n}=n $$ but since $\ker A = \ker B$ this implies that the images have the same dimension. This in turn implies that they're isomorphic: there is a linear bijection from $\operatorname{im} A$ to $\operatorname{im} B$. This implies the statement that there's $C$ such that $B=CA$: for any $v \in \mathbb{K}^n$ the vector $Av$ is in the image of $A$ and $Bv$ in the image of $v$. $C$ is now the representation matrix of the isomorphism taking $Ae_j$ to $Be_j$ for all vectors $e_j$ of a basis of $\mathbb{K}^n$ (note that you can construct this isomorphism without the rank-nullity theorem - I just think it helps to think about it beforehand).
And finally statement (A): Let $m=n=2$ and consider the matrices $$ A = \pmatrix{1 & 0 \\0 & 0} \quad B = \pmatrix{1 & 0 \\ 1 & 0}. $$ They both have the same kernel spanned by $e_2 := \pmatrix{0 \\ 1}$. Assume there's some matrix $C$ such that $B=AC$, then $$ AC=A \pmatrix{c_{11} & c_{12} \\ c_{21} & c_{22}} = \pmatrix{c_{11} & c_{12} \\ 0 & 0} $$ Let $e_1 = \pmatrix{1 \\ 0}$. We have $Be_1 = \pmatrix{1 \\ 1}$ but $ACe_1 = \pmatrix{c_{11} \\ 0}$. Since in all fields $1 \neq 0$ this contradicts that $B=AC$ so we found a counterexample to the claim.
EDIT: maybe it helps to consider how this counterexample works : consider a bit more easily understood the maps $(x,y) \mapsto (x,x)$ and $(x,y) \mapsto (x,0)$ on $\mathbb{R}^2$ (there correspond to $A$ and $B$). Clearly they both map all the points on the y-axis into $0$ - they just differ by what line they map the x-axis to. No transformation of $\mathbb{R}^2$ (this corresponds to $C$) can change this simple fact: it can't turn one of those lines into the other one because it "does its thing" before this transformation happens.
So abstractly this hinges on the images of $A,B$: while we have the same kernels we can still have different images and as long as that's the case we can't turn one into the other.