Two seemingly non-isomorphic elliptic curves over a finite field which have the same cardinality

Question

Let $p$ be a prime number such that $p \equiv 3 \bmod 4$. Now consider the elliptic curves $$ E/\mathbb{F}_{p^2}: \quad y^2 = x^3 - ax \quad \text{and} \quad E'/\mathbb{F}_{p^2}: \quad y^2 = x^3 - a^{-1}x $$ where $a$ is an element generating $\mathbb{F}_{p^2}^*$.

The curve itself depends on the choice of $a$: the curves $E$ and $E'$ are not isomorphic in general, see MAGMA code below (for my computed cases, they are all not isomorphic to each other). But what I noticed is that in every case I computed, it is $|E(\mathbb{F}_{p^2})| = |E'(\mathbb{F}_{p^2})|$.

Question: Is this a general observation or is it just mere coincidence?

Below is the MAGMA code I used:

for p in [ x : x in [3 .. 100] | IsPrime(x) and x mod 4 eq 3 ] do
    print "";
    K := GF(p^2);
    a := PrimitiveElement(K);
    R<x> := PolynomialRing(K);
    f := x^3 - a*x;
    g := x^3 - 1/a*x;
    E := EllipticCurve(f);
    Eprime := EllipticCurve(g);
    print "Are both curves isomorphic?";
    IsIsomorphic(E,Eprime);
    print "Do both curves have the same cardinality?";
    #E eq #Eprime;
end for;

Answer 1

Taking $c=a^{-(p+1)/4}\in \Bbb{F}_{p^2}$ $$E':y^2=x^3-a^{-1}x$$ is $\Bbb{F}_{p^2}$-isomorphic to $$(c^3y)^2= (c^2x)^3-a^{-1}(c^2 x)=c^6(x-a^{-1}c^{-4} x)$$ ie. to $$E^p:y^2 = x^3-a^p x$$ Whence $$|E'(\Bbb{F}_{p^2})|=|E^p(\Bbb{F}_{p^2})|=|E(\Bbb{F}_{p^2})^p|=|E(\Bbb{F}_{p^2})|$$ This also proves that the two curves are non-isomorphic.