Yes. If $Bx=0$, then $x^T B x = (Ax)^T (Ax) = \|Ax\|^2 = 0$. Since the columns of $A$ are linearly independent, $Ax=0$ iff $x=0$. Hence $x=0$ and the solution to $Bx=0$ is not just unique, it is zero.
Yes. If $Bx=0$, then $x^T B x = (Ax)^T (Ax) = \|Ax\|^2 = 0$. Since the columns of $A$ are linearly independent, $Ax=0$ iff $x=0$. Hence $x=0$ and the solution to $Bx=0$ is not just unique, it is zero.