= Solution
The squares in $\mathbb F_5$ are $0,1,4$. The following table counts affine solutions at each $x$, with one point at infinity to be added:
$$
\begin{array}{c|ccccc}
x&0&1&2&3&4\\\hline
x^3-x&0&0&1&4&0\\
\#\{y:y^2=x^3-x\}&1&1&2&2&1\\
x^3-x+1&1&1&2&0&1\\
\#\{y:y^2=x^3-x+1\}&2&2&0&1&2
\end{array}
$$
Hence $\#E_1(\mathbb F_5)=\#E_2(\mathbb F_5)=8$. In characteristic five, a finite point has order two precisely when $y=0$. The first cubic has three roots $0,1,4$, giving four elements killed by two, including $O$. The second cubic has only the root $3$, giving two elements killed by two. An isomorphism of <groups> would preserve these counts, so the <groups> are not isomorphic.
More precisely $(2,1)\in E_1$ doubles to $(0,0)$ and has order four; the independent two-torsion point $(1,0)$ gives $C_4\times C_2$. On $E_2$, the point $(0,1)$ doubles to $(4,1)$, which doubles to $(3,0)$; it therefore has order eight. Thus the <eight-point isogenous elliptic curves over F5> have point <groups>
$$
\boxed{E_1(\mathbb F_5)\cong C_4\times C_2,\qquad E_2(\mathbb F_5)\cong C_8.}
$$
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