According to Derrick's theorem we can write \begin{align} E &= \frac{1}{2} \int d^Dx \frac{1}{\lambda^2}\left( \nabla \phi_i (\frac{x}{\lambda})\right)^2 + \int d^Dx V(\phi_i(\frac{x}{\lambda})),\\ &=\lambda^{D-2} I_K +\lambda^{D} I_V. \tag{1} \end{align} $\lambda = 1$ must be a stationary point of $E(\lambda)$, which implies that, \begin{equation} 0=(D-2) I_K[\bar\phi]+D I_V[\bar\phi]. \end{equation} I don't understand is
For $D \geq 3$, ($1$) can only be satisfied if both $I_K$ and $I_V$ vanish, which rules out anything but a constant vacuum solution.
Can you elaborate this please.
I'm not sure about the origin of all the terms, but I think what's happening is that both $I_K$ and $I_V$ are positive. This means that for the expression
$$\begin{equation} 0=(D-2) I_K[\bar\phi]+D I_V[\bar\phi]. \end{equation}$$
to work out, $D-2$ has to be negative. If $D-2>0$, then you'll only get zero if both $I_K$ and $I_V$ are zero.