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 $x\mapsto x-1$ 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 $E_{2,1-t}$.

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 $E_{2,1-t}$.