Degree of morphisms and isogenies $\renewcommand{\J}{\mathrm{Jac}} \renewcommand{\F}{\mathbb{F}}$
I am reading this paper by B. Gross, and there is something I don't understand on p. 945. Here is the context: fix a prime $p \equiv 3 \pmod 4$, and define the (hyper)elliptic curves over $\F_p$ given by the (affine) equations
$$X_1 : y^2 = x^p - x,\quad X_2 : y^2 = x^{p+1}-1,\quad 
  E_1 : y^2 = x^3-x,\quad   E_2 : y^2=x^4-1.$$
I checked (using Tate isogeny theorem) that there is a non-zero isogeny $\alpha : \J(X_1) \to \J(X_2)$ between the jacobian varieties (actually both are isogenous to $E_1^{(p-1)/2}$ over $\F_p$), and there is a non-zero isogeny $\beta : E_1 \to E_2$.

Question: There is a morphism $f_2 : X_2 \to E_2, (x,y) \mapsto (x^{(p+1)/4}, y)$ which has degree $(p+1)/4$. Then it is claimed that we therefore get a morphism $f_1 : X_1 \to E_1$ of degree $(p+1)/4$, but I don't see why/how.


Thoughts:
I know that $f_1$ induces a morphism $\phi_2 : \J(X_2) \to E_2$, we get a morphism $\beta \circ \phi_2 \circ \alpha^{\vee} : \J(X_1) \to E_1$, hence a morphism $f_1 : X_1 \to E_1$, but I believe that it has degree at least the degree of $f_2$. 
Maybe there is a clever way to compose $\phi_2$ with other isogenies to get the equality of degrees?
In general, given a non-constant morphism $f_2 : X_2 \to E_2$, it might not be possible to get a morphism $f_1 : X_1 \to E_1$ of same degree as $f_2$: just take $X_2 = E_2 = X_1, f_2 = \mathrm{id}$ and $E_1$ an elliptic curve isogenous but not isomorphic to $E_2$.
I am probably missing something easy, but I prefer asking for clarifications.
 A: It's easier if we forget about isogenies: $E_1$ and $E_2$ are isomorphic,
and $X_1$ and $X_2$ are isomorphic, so the cover $X_2\to E_2$
induces a cover $X_1 \to E_1$ of the same degree. 
To make this more explicit:
let $d = (p+1)/4$,
and rewrite the curve equations in separate coordinate systems:
\begin{align*}
    X_1: Y^2 & = X^p - X \,,
    &
    X_2: V^2 & = U^{4d} - 1 \,,
    \\
    E_1: y^2 & = x^3 - x \,, 
    &
    E_2: v^2 & = u^4 - 1 \,.
\end{align*}
Let $i$ be a square root of $-1$ in $\mathbb{F}_{p^2}$.
Then there is an isomorphism $\psi: X_1 \to X_2$
defined by
$$
    \psi: (X,Y)
    \longmapsto
    (U,V) = \left(
        \frac{X + i}{X - i}
        ,
        \frac{(i+1)Y}{(X -i)^{2d}}
    \right)
$$
(here we need $i^p = -i$),
the degree-$d$ cover $f_2: X_2 \to E_2$ defined by
$$
    f_2: (U,V) \longmapsto (u,v) = (U^d, V)
    \,,
$$
and an isomorphism $\phi: E_2 \to E_1$
defined by
$$
    \phi: (u,v)
    \longmapsto
    (x,y) = \left(
        -i\cdot\frac{u + i}{u - i}
        ,
        \frac{(i+1)v}{(u-i)^2}
    \right)
    \,.
$$
Composing,
we get a degree-$d$ cover $f_1 = \phi\circ f_2\circ\psi: X_1 \to E_1$,
which is what we wanted...  Well, almost what we wanted, because we would probably want $f_1$ to be defined over $\mathbb{F}_p$. 
But expanding, we see that $f_1$ is defined by
$$
    f_1: (X,Y)
    \longmapsto
    (x,y)
    =
    \left(
            -i\cdot\frac{
                (X + i)^d + i(X-i)^d
            }{
                (X + i)^d - i(X-i)^d
            }
            ,
            \frac{ 2i Y }{
                ((X + i)^d - i(X-i)^d)^2
            }
    \right)
    \,,
$$
Both of the rational functions
are symmetric with respect to $i \leftrightarrow -i$,
so they are defined over $\mathbb{F}_p$,
and therefore so is $f_1$.
All four curves have plenty of automorphisms (some over $\mathbb{F}_p$, some over $\mathbb{F}_{p^2}$) which you can compose with these morphisms to produce more solutions.
