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 $$ \psi: (X,Y) \longmapsto (U,V) = \left( i\cdot\frac{X + i}{X - i} , \frac{(i+1)Y}{(X -i)^{2d}} \right) $$ (here we need $i^p = -i = 1/p$), 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 $$ \phi: (u,v) \longmapsto (x,y) = \left( i\cdot\frac{u + i}{u - i} , (-1)^{d}\cdot\frac{(i-1)v}{(u-i)^2} \right) \,. $$ 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 $$ f_1: (X,Y) \longmapsto (x,y) = \left( i\cdot\frac{ i^d(X + i)^d + i(X-i)^d }{ i^d(X + i)^d - i(X-i)^d } , \frac{ (-1)^{d+1}2 Y }{ (i^d(X + i)^d - i(X-i)^d)^2 } \right) \,, $$ 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 $f_1$.

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.