This might be a counterexample to a conjecture of Granville about automorphisms of twists of hyperelliptic curves.
In this paper,
the quadratic twist of $f(x)=y^2$ is denoted by $C_d : d y^2=f(x)$ on p.1. where $f(x)$ is squarefree.
From p. 3 We denote by $Aut(C)$ the (finite) group of such automorphisms of $C$, and define $c_d(\mathbb{Z})$ and $c_d (\mathbb{Q})$ to be the number of automorphism classes of non-trivial integral and rational points on $C_d$ , respectively.
Conjecture 2 (ii) There are only finitely many squarefree integers $d$ for which $c_d (\mathbb{Q}) > 2$ when $\deg(f)>d_0$.
Proposed counterexample which stands numerical experiments.
Let $a(x)=(-x^2+5x+1)^2,b(x)=(-x^2-5x+1)^2$ and $f(x)=a(x)^n+b(x)^n$ for even natural $n$.
$f(x)$ is even and symmetric, so we have the automorphisms $(x,y) \mapsto (-x,y), (x,y) \mapsto (1/x,1/x^{2n})$ for all $d$.
Let's solve $a(x_1)=A,b(x_1)=B,a(x_2)=Bu,b(x_2)=Au$ over the rationals.
Let $D=numerator(A^n+B^n)$. Write $D=d D'^2$ with $d$ squarefree. We have $f(x_1),f(x_2)$ on $C_d$ since $f(x_1)=D=d D'^2$ and $f(x_2)=Du^n=d u'^n$ ($n$ is even).
Solving symbolically gives three solutions. The first two are the known automorphisms and the third is
$x_2=1/2/(-1+x_1^2)(-25x_1+(y_1 ))$ for $x_1,y_1$ on $C : 617x_1^2+4x_1^4+4-y_1^2=0$
$C$ has infinitely many rational points, which gives the third solution on $C_d$ whenever $x_1,y_1$ are on $C$.
So third candidate automorphism is $(x_1,y) \mapsto (1/2/(-1+x_1^2)(-25x_1+(y_1 )),\ldots)$.
To show the candidate is automorphism, one must show $x_2$ is $x_1$ coordinate on $C$, which is experimentally true (and this is independent of $n$).
We tried to showed the above, but didn't succeed.
Q1 Can this argument be made rigorous?
As an aside, this gives infinitely many rational and integer non-trivial solutions to $(a(x_1)^n+b(x_1)^n)(a(x_2)^n+b(x_2)^n)=y^2$.
Sage verification, code, might be run in the cloud:
def granvsym2():
K.<x>=QQ[]
#on 617*x1^2+4*x1^4+4-y1^2=0
n=8 #change to arbitrary large even
E=EllipticCurve(QQ,[0, 0, 0, -380881/3, 469059442/27])
P=E.gens()[0]
h=(-x^2-5*x+1)^(2*n)
f=(-x^2+5*x+1)^(2*n) + h
g=(617*x^2+4*x^4+4)
for k in [ 2 .. 10]:
u,v=(k*P).xy()
u= -u
x1,y1=-36/(380113+3702*u+9*u^2)*v, -1/2*(-7611476-29616*u+36*u^2)/(380113+3702*u+9*u^2)
x2=1/2/(-1+x1^2)*(-25*x1+(y1 ))
print k,QQ(f(x1)/f(x2)).is_square(),g(x1)-y1^2==0,'---','x2squ=',QQ(g(x2)).is_square()