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Let $f(x,y) \in \mathcal{O}^\ast_{\mathbb{C}^2,0}$, a germ of holomorphic function at the origin of $\mathbb{C}^2$ with $f(0,0)=1$. Let $$\varphi(x,y)=(ax+by,cx+dy)$$ be a linear germ of biholomorphism ( $ad-bc\neq0$) of finite order, $\varphi^n(x,y)=(x,y)$. Suppose that $$\prod_{i=0}^{n-1}f\circ\varphi^i =1.$$ Is it true that $f=1$? I think this shuold not be a hard question but I could not figure out how to prove or find a counterexample.

EDIT: It is false, as seen in the comments. However, is there any known structure for such functions for a given $\varphi$?

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    $\begingroup$ Say, $f(x,y)=\frac{1-x}{1-y}=(1-x)(1+y+y^2+...)$, with $\varphi(x,y)=(y,x)$? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 18, 2017 at 8:20
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    $\begingroup$ Conjugating, one can reduce to the cases $\varphi_k(x,y):=(e^{\frac{2\pi i}n}x,e^{\frac{2\pi i k}n}y)$. It should not be difficult to compute all possible cases, I think they form (up to invertible changes of variables) a finite-dimensional vector space for each $n$. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 19, 2017 at 12:41

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