Prove $\{f(x,y) \in \mathbb{C}[[x,y]] \mid f(\zeta_n x,\zeta_n^{-1}y) = f(x,y)\}$ is not isomorphic to the formal series ring 
Suppose that $\zeta_n$ is a primitive n-th root of $1$.
Let $R$ be $\{f(x,y) \in \mathbb{C}[[x,y]] \mid f(\zeta_n x,\zeta_n^{-1}y) = f(x,y)\}$.
Try to prove that $R$ is not isomorphic to $\mathbb{C}[[x,y]]$ as $\mathbb{C}$-algebras.

My Work: Assume that $f(x,y)=\sum a_{i,j}x^iy^j$. Then $f(\zeta_n x,\zeta_n^{-1}y)=f(x,y)$ implies that $a_{i,j}=0$ when $n\not\mid i-j$. Thus $$f(x,y)=\sum_{in+j\geq 0}(xy)^jx^{in}.$$ Then I can't go any further. I think we need some properties of $\mathbb{C}[[x,y]]$.
Any one has some advice or ideas?
And there is another question, let $S=\{f[x,y]\in \mathbb{C}[[x,y]]\mid f[\zeta_nx,\zeta_ny]=f[x,y]\}$. Try to prove that $S$ is not isomorphic to $R$ as $\mathbb{C}$-algebras.

Each of the three rings $\mathbb{C}[[x,y]]$, $R$, and $S$ is local, with a unique maximal ideal consisting of power series with constant term $0$. For each ring, I will compute the vector space dimension of the quotient of the ring by the square of the maximal ideal. The three dimensions will be distinct, which proves the three rings are pairwise non-isomorphic as $\mathbb{C}$-algebras.
For $\mathbb{C}[[x,y]]$, the maximal ideal $\mathfrak{m}$ is the set of power series whose monomials $x^iy^j$ satisfy $i+j\geq 1$, so $\mathfrak{m}^2$ is power series whose monomials $x^iy^j$ satisfy $i+j\geq 2$. The quotient $\mathbb{C}[[x,y]]/\mathfrak{m}^2$ has a basis $1$, $x$, $y$, so is $3$-dimensional.
The ring $R$ consists of power series whose monomials $x^iy^j$ satisfy $n|i-j$. The maximal ideal $\mathfrak{m}$ is the set of power series whose monomials $x^iy^j$ satisfy $i+j\geq 1$ and $n|i-j$. By taking pairwise products of such monomials, we can get monomials $x^iy^j$ satisfying either $i$, $j\geq 2$, or $i\geq 1$, $j\geq n$, or $j\geq 1$, $i\geq n$. The quotient $R/\mathfrak{m}^2$ has a basis $1$, $xy$, $x^n$, $y^n$, so is $4$-dimensional.
The ring $S$ consists of power series whose monomials $x^iy^j$ satisfy $n|i+j$. The maximal ideal $\mathfrak{m}$ is the set of power series whose monomials $x^iy^j$ satisfy $n|i+j$ and $i+j\geq n$, so $\mathfrak{m}^2$ is the set of power series whose monomials $x^iy^j$ satisfy $n|i+j$ and $i+j\geq 2n$. The quotient $S/\mathfrak{m}^2$ has a basis 
$$
1,x^n, x^{n-1}y, x^{n-2}y^2,\ldots,xy^{n-1},y^n,
$$
so is $(n+2)$-dimensional.
