When playing with the scaling problem
$$G(4z)=\frac{G(z)}{2z}$$
(see also this question) I discovered, that the general problem
$$G(\lambda z)=\frac{G(z)}{\gamma z}$$ with two constants $\lambda,\gamma>0$ can be solved by
$$G(z)=kz^{\displaystyle{a+b\ln z}},\qquad a=\frac{1}{2}-\frac{\ln\gamma}{\ln\lambda},\qquad b=\frac{-1}{2\ln\lambda}.$$
But how would one approach the more general problem
$$G(\lambda z)=\frac{G(z)}{\gamma z^n}$$ with a free power $n\in\mathbb{N}$?
Let $z=\lambda^u$ ,
Then $G(\lambda\lambda^u)=\dfrac{G(\lambda^u)}{\gamma(\lambda^u)^n}$
$G(\lambda^{u+1})=\dfrac{G(\lambda^u)}{\gamma\lambda^{nu}}$
$G(\lambda^u)=\prod\limits_u\dfrac{1}{\gamma\lambda^{nu}}$
$G(\lambda^u)=\gamma^{-u}\lambda^{-n\sum\limits_uu}$
$G(\lambda^u)=\Theta(u)\gamma^{-u}\lambda^{-\frac{nu(u-1)}{2}}$ , where $\Theta(u)$ is an arbitrary periodic function with unit period
$G(z)=\Theta(\log_\lambda z)\gamma^{-\log_\lambda z}\lambda^{-\frac{n(\log_\lambda z-1)\log_\lambda z}{2}}$ , where $\Theta(z)$ is an arbitrary periodic function with unit period