TITLE: An isogeny between Jacobians of hyperelliptic curves
QUESTION [12 upvotes]: Let $\mathbf{F}_q$ be a finite field of odd characteristic.  Let $X_t$ be the hyperelliptic curve over $\mathbf{F}_{q^2}(t)$ with affine equation
$$y^2 = \left((x^{(q+1)/2}-(x-1)^{(q+1)/2})^2 - t\right)
\left((x^{(q+1)/2}+(x-1)^{(q+1)/2})^2 - t\right).$$
This family of hyperelliptic curves is smooth of genus $q-1$ away from $t=0,1,\infty$.  For theoretical reasons I expect the following claim to be true:

The Jacobians of $X_t$ and $X_{1-t}$ are isogenous.  Equivalently, there is a nontrivial correspondence between $X_t$ and $X_{1-t}$.

It is easy to verify (with MAGMA for instance) that the zeta functions of $X_t$ and $X_{1-t}$ agree for small values of $q$ and specific values of $t$, to the point that I'm quite convinced of the claim.  To prove it, one could try to construct the correspondence explicitly.  There is some literature about pairs of hyperelliptic curves with isogenous Jacobians (for instance this paper of Mestre), but it seems to involve finding a congruence between bivariate polynomials, which comes out of nowhere.  
Short of constructing the actual correspondence, is there an algorithm for deciding whether two hyperelliptic curves admit an isogeny of given degree between their Jacobians?

REPLY [4 votes]: In the most easy case $q=3$, the curve $X_t$ is bielliptic (the bielliptic involution given by $x\mapsto 1-x$), and the Jacobian of $X_t$ is then $(2,2)$-isogenous to the product 
$E_{1,t}\times E_{2,t}$, where $E_{1,t}$ and $E_{2,t}$ are the elliptic curves 
$$E_{1,t}: y^2=x(x^2 - (t+1)x + (t- 1)^2)$$
and 
$$E_{2,t}: y^2=x(x^2 + x - (t^3 - 1))$$
Now, $E_{1,t}$ is 2-isogenous to 
$$E_{1,t}': y^2= x(x+1)(x-(t-1))$$
and one has that 
$$E_{1,t}'\cong E_{1,1-t}$$
by sending $x\mapsto 1-x$ and then twisting by $-1$, which is an isomorphic curve over $\mathbb{F}_9$ (this is the only place where we use we are over $\mathbb{F}_9$ and not over $\mathbb{F}_3$).
while $E_{2,t}$  is 2-isogenous to 
$$E_{2,t}': y^2 = x(x^2 + x + t^3)$$
which in turn is clearly equal to $E_{2,1-t}$.