Let hyper elliptic curve $C$ be given by equation $C: y^2=x^6+2x^5+x^4-6x^3+2x^2-4x+1$.
It is known by general theory that every autmorphism of $C$ is given by $φ: (x,y)\mapsto(\frac{ax+b}{cx+d},\frac{(ad-bc)y}{(cx+d)^3})$, $a,b,c,d \in\overline {\Bbb{Q}}$.
I want to find an endmorphism $φ$ of $C$ which satisfies $φ^2+φ-1=0$. So what I should do is just to find $(a,b,c,d)$. But Equation of $C$ is not factorable anymore, so straightforward calculation seems almost impossible for human. Using computer or result only is welcomed, thank you for your help.
P.S. I posted very similar question with typos. Sorry and thank you so much for people comment or answer to that, but the typos was too severe, I reposted this question.
Let $\phi$ be the map $$(x,y)\mapsto \left(\frac{ax+b}{cx+d},\frac{(ad-bc)y}{(cx+d)^3}\right)$$ and assume that it is an automorphism of $C$ satisfying $\phi^2+\phi-{\rm id}=0$. This quadratic condition immediately implies that $b=c=0$, and then the computation is very easy (and "possible for human"), and gives \begin{align*} 0 & = a^2 + ad - d^2,\\ 0 & = a^2 + ad^2 - d^4, \end{align*} with $a,d\neq 0$. The solutions of these two equations are given by either $a^2+a-1=0$ and $d=1$, or $a=d+d^2$ and $d^2 + 3d + 1=0$.
Edit: Now we can check explicitly whether or not these four maps are automorphisms of $C$ or not.