= Solution
Let $\pi,\pi'$ be the $p$-power <Frobenius isogenies> of $E,E'$. Since $\psi$ is defined over $\mathbb F_p$, it commutes with Frobenius:
$$
\psi\pi=\pi'\psi,\qquad\psi(1-\pi)=(1-\pi')\psi.
$$
Taking degrees, using multiplicativity and cancelling the nonzero <degree of an isogeny> $\deg\psi$, gives
$$
\deg(1-\pi)=\deg(1-\pi').
$$
Both differences are separable, so the preceding proof identifies their degrees with rational point counts. Therefore
$$
\boxed{\#E(\mathbb F_p)=\#E'(\mathbb F_p).}
$$
<Isogenous elliptic curves can have different rational point groups>. Here is an explicit example. Over $\mathbb F_7$, take
$$
E:y^2=x^3-x,\qquad E':Y^2=X^3+4X.
$$
The map
$$
\psi(x,y)=\left(x-\frac1x,\ y\left(1+\frac1{x^2}\right)\right)
$$
extends across $O$ and $(0,0)$ to a degree-$2$ <isogeny of elliptic curves> with those two points as its kernel. Substitution verifies the target equation, or this follows from the <two-isogeny formula> with $a=0,b=-1$. Both curves are smooth modulo $7$.
For $x=0,1,\ldots,6$, the numbers of affine points with that abscissa on $E$ are $(1,1,0,0,2,2,1)$, and on $E'$ they are $(1,0,2,2,0,0,2)$. Adding the point at infinity gives $8$ on each. The first curve has four rational points of order dividing $2$, from $O$ and the three roots $0,1,-1$. The second has only two: its quadratic factor $X^2+4$ has no root modulo $7$, since $3$ is not a square. By the classification of <finite abelian groups>,
$$
\boxed{E(\mathbb F_7)\cong\mathbb Z/2\mathbb Z\times\mathbb Z/4\mathbb Z,\qquad E'(\mathbb F_7)\cong\mathbb Z/8\mathbb Z.}
$$
They are thus isogenous with equal orders and nonisomorphic groups.
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