mathematical physics - functional equation in renormalization group theory

In the Renormalization Group Theory, a key step is the derivation of the so called scaling equation, which in general is in the form of a functional equation of the kind:
$$g(\mu(\lambda)x,\nu(\lambda)y) = \lambda g(x,y)$$

where x, y are real numbers and $\lambda>0$, moreover $\mu(\lambda=1)=\nu(\lambda=1)=1$.

We can assume g differentiable both on x and y and $\mu(\lambda),\nu(\lambda)$ analytic function of $\lambda$.

The (hopefully unique) solution of the functional equation should correspond to $\mu(\lambda)$ and $\nu(\lambda)$ being of the form of power law with two independent exponents.

Is it necessary to add other conditions on the functions $g,\mu$ and $\nu$ to prove this result and its uniqueness? Or is it possible to weaken the hypotheses?

Edit:

Assuming $\mu(\lambda)$ being an invertible function (this is an additional hypothesis), setting $y=0$ we get
$$g(\lambda x,0) = \mu^{-1}(\lambda) g(x,0)$$
It easy to get
$$g(\lambda_1 \lambda_2 x,0) = \mu^{-1}(\lambda_1)\mu^{-1}(\lambda_2) g(x,0)= \mu^{-1}(\lambda_1 \lambda_2) g(x,0)$$
from which follows (at least for differentiable $\mu^{-1}(\lambda$):

$$\mu^{-1}(\lambda) = \lambda^s$$ and then $\mu(\lambda) = \lambda^{\frac{1}{s}}$.

My question is if it is necessary to add an additional hypothesis like that of an invertible $\mu(\lambda)$, if there are alternatives to the invertibility or if is possible to proof the result even under weaker conditions.