We define the energy as $$E = I_F + I_K + I_V,$$ where, $$I_F [A]= \frac{1}{2} \int d^Dx \operatorname{tr} F^2_{ij},$$ $F_{ij}$ represents the electromagnetic force. $$I_K [\phi,A]= \frac{1}{2} \int d^Dx (D_j \phi)^\dagger (D_j \phi),$$ $$I_V [\phi]= \int d^Dx V(\phi),$$ Given a solution $\bar \phi(x)$, $\bar A_j(x)$, we define scaled fields $$f_\lambda(x)= \bar\phi(\lambda x),$$ $$g_{j\lambda}(x)= \lambda \bar A_j(\lambda x).$$ Proceeding as in the pure scalar case, we find that \begin{align} E(\lambda) &= I_F [g_\lambda ] + I_K [f_\lambda , g_\lambda ] + I_V [f_\lambda ] \\ & = \lambda^{4-D} I_F [\bar A] + \lambda^{2-D} I_K [\bar \phi,\bar A]+ \lambda^{-D} I_V [\bar \phi] \tag{1} \end{align} which is stationary at $\lambda = 1$ if
I just don't understand how $$I_F [A]=\lambda^{4-D} I_F [\bar A],$$ is in the equation (1)