Linear automorphisms of $y^3 + y^6 = x^6$ I have the curve $C \subset \mathbb{C}^2$ defined by the equation $y^3 + y^6 = x^6$ and I have to prove that the set of maps of the form $\varphi = (f_1,f_2) \in \operatorname{Aut}(\mathbb{C}^2)$ such that $\deg(f_1) = \deg(f_2) = 1$ and $\varphi (C) = C$ is a group isomophic to $\mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$.
It's clear that $\left\{ \varphi (x,y) = \left(e^{\frac{2\pi i}{6} k} x,e^{\frac{2\pi i}{3} k'} y \right): 1 \leqslant k \leqslant 6,1 \leqslant k' \leqslant 3\right\}$ holds all conditions and it's isomophic to $\mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$, but I can't prove that these morphisms are all.
Can somebody give me a guidance?
 A: Using Jacobian criterion, we see that $(0,0)$ is the only singular point of $C$. As automorphisms preserve the singular locus, this forces $\varphi(0,0)=(0,0)$, hence $\varphi$ is actually linear. 
As $f_2^3+f_2^6-f_1^6=0$ on $C$ and the ring of regular functions on $C$ is $\mathbb C[X,Y]/(Y^3+Y^6-X^6)$, we get 
$$f_2(X,Y)^3+f_2(X,Y)^6-f_1(X,Y)^6=c(Y^3+Y^6-X^6)$$
for some constant $c\in \mathbb C^*$. Write $f_1(X,Y)=a_{11}X+a_{12}Y, f_2(X,Y)=a_{21}X+a_{22}Y$. Substituting $Y$ with $0$ then gives
$$a_{21}^3X^3+a_{21}^6X^6-a_{11}^6X^6=-cX^6.$$ 
Hence $a_{21}=0$ and $f_2(X,Y)=a_{22}Y$. Therefore
$$a_{22}^3Y^3+a_{22}^6Y^6-(a_{11}X+a_{12}Y)^6=c(Y^3+Y^6-X^6).$$ 
This immediately implies that $a_{12}=0$, and $a_{22}^3=c=a_{22}^6$, $a_{11}^6=c$. Hence $a_{22}^3=a_{22}^6/a_{22}^3=c/c=1$ and $c=a_{22}^3=1$, thus $a_{11}^6=1$. 
Conclusion: $f_1(X,Y)=aX$ with $a^6=1$, $f_2(X,Y)=bY$ with $b^3=1$ and $\varphi(x,y)=(ax, by)$. This proves there is no other automorphisms than that you already found. 
A: As QiL observes, an affine automorphism $\alpha$ of the plane which preserves the curve must preserve the singular point, and therefore it must be linear.
There is only one line through the singular point which has order of contact there equal to $6$, namely the line $L$ with equation $y=0$, so this line must be preserved by any linear automorphism.
Moreover, the automorphism maps every line parallel to $L$ to a line parallel to $L$. There are exactly three lines parallel to $L$ which intersect the curve in only one point, and these determines three points and it happens that they are on a line $R$. The line $R$ must be preserved by the automorphism, too. One easily sees that $R$ has equation $x=0$.
It follows that $\alpha$ is not only linear but in fact it is diagonal.
Using this piece of information, one finds all possibilities easily.
A simple way of this this is the following. Suppose $\alpha(x,y)=(ax,by)$; of course $ab\neq0$.  Then whenever $y^3+y^6=x^6$ we have $b^3y^3+b^6y^6=a^6x^6$ and therefore $$(a^6-b^6)y^6+(a^6-b^3)y^3=0. \tag{$\star$}$$ This last equation holds for every $y$ in the field such that there is an $x$ with $(x,y)$ in the curve; since the field is algebraically closed, this condition in fact always holds, and we see that the equality $(\star)$ holds for all $y$. This is only possible if $a^6=b^6=b^3$. It follows at once from this that $b$ is a cubic root of unity and $a$ a sixth root of unity.
