Consider the projective $2$-space $\mathbb{P}^2$ over the field $\mathbb Q$ given by the homogeneous cordinates $[X:Y:Z]$. Cinsider the following two elliptic curves in homogeneous coordinates: $$E_1: (X^3+Y^3=Z^3)~~\text{and}~~E_2:=(Y^2Z=X^3-432Z^3).$$ We can think of the solution sets of $E_1,E_2$ in $\mathbb{P}^2$ as projective subvarieties $V(E_1)$ and $V(E_2)$ respectively.
Then the rational map $\varphi:=[12Z, 36(X-Y),X+Y]$ is a morphism from $V(E_1) \to V(E_2)$, since $\varphi$ is defined at every point of $\mathbb P^2(\mathbb Q)$.
How to show $\varphi$ is an isomorphism ?
We know that $\varphi$ will be isomorphism if there is another morphism $\psi: V(E_2) \to V(E_1)$ such that $$\varphi \circ \psi =id_{V(E_2)},~~\psi \circ \varphi=id_{V(E_1)}.$$
So let $\psi:=[f_1,f_2,f_3]$ be another morphism such that $$\varphi \circ \psi=id_{V(E_2)}=[1,1,1],$$ where $f_i, ~i=1,2,3$ are homogeneous polynomials in $\mathbb Q[X,Y,Z]$.
If I am correct, then the last equation gives $$ f_1=Y/72+Z/2, f_2=Z/2-Y/72, f_3=X/12$$ i.e., $$\psi:=[Y/72+Z/2, Z/2-Y/72, X/12],$$ which exists at all points of $\mathbb P^2(\mathbb Q)$.