How is $f$ restricted if we require that $f(A\cdot x)=g(A)f(x)+h(A)$?

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EDIT: I see now that there was a mistake in my latex, causing me to ask a completely different and more trivial quesiton than I intended. I have edited the question:

Question:

How is the functional form of $f(x)$ restricted if we require that for any $A$ and any $x$, there exists a $g(x)$ and $h(x)$ such that $f(A\cdot x)=g(A)f(x)+h(A)$?

Also, if the $h$ over complicates things, I'm also interested in the question if we assume $h(x)=0$

Motivation:

If I have a general equation: $$Af(x)=f(y)$$ and I would like to transform this into: $$g(A)x +h(A)=y$$

By taking the inverse of $f$: $$f^{-1}(Af(x))=f^{-1}(f(y))$$ and then assuming that we can "take out $A$" out of the inverse (The desired property is thus a property of $f^{-1}$): $$g(A)\cdot f^{-1}(f(x))=f^{-1}(f(y))$$ So that we find a linear relation between $x$ and $y$. $$g(A)x=y$$

I'm wondering what kind of restrictions this puts on $f$.

  1. I know that this would work if $f$ is homogenous, since then also $f^{-1}$ is homogenous so that we can write: $f^{-1}(Af(x))=A^\lambda x$

  2. But I'm not sure if there are any other functional forms that satisfy the desired property.

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The identity $$Af(x) = xg(A)+h(A)$$ implies $$f(x) = xg(1)+h(1)$$ by taking $A=1$, so that the identity only holds for $f$ an affine function, with $h(A)\equiv f(0)$ and $h(A)\equiv f'(0)$.