 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: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