If matrices $A$ and $B$ have the same fundamental subspace than $A=cB$ where $c$ is some scalar.
I think it is true, because I try to disprove with this two matrices, they have the same dimension of fundamental subspace, but I think in this question want that we have the same vector in fundamental space of matrices. So than is true that exist c.
A=$\begin{bmatrix} 1& 0& 0\\ 0& 1& 0\\ 0 & 0& 1\\ 0 & 0& 2 \end{bmatrix}$. B=$\begin{bmatrix} 3& 0& 0\\ 0& 2& 0\\ 0 & 0& 1\\ 0 & 0& 2 \end{bmatrix}$.
But what do you think?
If by fundamental subspace you mean the space generated by rows and the kernel of $A$ and $A^t$ as stated here then your assertion is false. Consider $$ A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \qquad B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} $$ Those two have the same fundamental spaces but there exists no $c \in \mathbb{R}$ such that $A = cB$