I have $\mathbf{x}(t) = \mathbf{V} \mathbf{u}(t)$.
Now I have to show that $\mathbf{x}'(t) = \mathbf{A} \mathbf{x}(t)$ implies that $\mathbf{u}'(t) = \mathbf{D} \mathbf{u}(t)$.
How do I do this?
I think I should use that $$ \mathbf{x}(t) = \mathbf{V} \mathbf{u}(t) \Leftrightarrow \mathbf{u}(t) = \mathbf{V}^{-1} \mathbf{x}(t) $$ and that the diagonal matrix $\mathbf{D}$ is defined as $$ \mathbf{D} = \mathbf{V}^{-1} \mathbf{A} \mathbf{V} \Leftrightarrow \mathbf{V}^{-1} = \mathbf{D} \mathbf{V}^{-1} \mathbf{A}^{-1}. $$ and insert $\mathbf{V}^{-1}$ in $\mathbf{u}(t)$.
You are right: by using $x(t) = V u(t)$ ($V$ constant matrix) and the fact that $x(t)$ satisfies the differential equatin $x'(t) = A x(t)$, you have
\begin{equation*} Vu'(t) = x'(t) = A x(t) = AV u(t) \end{equation*} hence \begin{equation*} u'(t) = V^{-1}AV u(t) \end{equation*}
Since the diagonal matrix $D$ is given by $V^{-1} A V$ you are done.