If $X(t)$ and $Y(t)$ are fundamental matrices then exists $C\in M_n$ invertible such thta $Y(t)=X(t)C$ Both for the homogeneous linear system $x'=A(t)x$.
attempt: Since $X$ is fundamental, it spans the solution space i.e., $$X'(t)=A(t)X(t)$$ but doing the same for $Y$ I go nowhere on finding $C$ invertible. I now that since $X$ is fundamental there is $t_0$ such that $\det(X(t_0))\neq 0$, even more $\det(X(t))\neq 0$ for all $t$, but again go nowhere.
You determine $C=X(t_0)^{-1}Y(t_0)$ at one point and then use that both solve the same linear differential equation to confirm that this relation holds with a constant $C$, that is, outright define $Y(t)=X(t)C$ and conclude from $$ X'=AX\\ (XC)'=X'C=AXC\\ Y'=AY $$ that $Y$ is indeed a solution.