Suppose $E_1$ and $E_2$ are elliptic curves defined over $\mathbb{Q}$.
Now we know that both curves are isomorphic over $\mathbb{C}$ iff
they have the same $j$-invariant.
But $E_1$ and $E_2$ could also be isomorphic over a subfield of $\mathbb{C}$.
As is the case for $E$ and its quadratic twist $E_d$. Now the question general is.
$E_1$ and $E_2$ defined over $\mathbb{Q}$ and isomorphic over $\mathbb{C}$. Let $K$
the smallest subfield of $\mathbb{C}$ such that $E_1$ and $E_2$ become isomorphic over $K$.
What can be said about $K$. Is it always a finite extension of $\mathbb{Q}$. If so, what can be
said about the extension $K|\mathbb{Q}$.
My second question is something goes something like in the opposite direction. I start again with
quadratic twists. Let $E$ be an elliptic curve over $\mathbb{Q}$ and consider the quadratic extension
$\mathbb{Q}|\mathbb{Q}(\sqrt{d})$. Describe the curves over $\mathbb{Q}$(or isomorphism classes over $\mathbb{Q}$)
which become isomorphic to $E$ over $\mathbb{Q}(\sqrt{d})$. I think the answer is $E$ and $E_d$.
Again I would like to know what happens if we take a larger extension.
Let $E$ be an elliptic curve over $\mathbb{Q}$ and $K|\mathbb{Q}$ a finite extension.
Describe the isomorphism classes of elliptic curves over $\mathbb{Q}$ which become isomorphic
to $E$ over K.
I have no idea what is the right context to answer such questions.
Best Answer
Question 1: Putting both curves in say, Legendre Normal Form (or else appealing the lefschetz principle) shows that if the two curves are isomorphic over $\mathbf{C}$ then they are isomorphic over $\overline{\mathbf{Q}}$. Now we could say that for instance $E_2$ is an element of $H^1(G_{\overline{Q}}, Isom(E_1))$ where we let $Isom(E_1)$ be the group of isomorphisms of $E_1$ as a curve over $\mathbf{Q}$ (as in Silverman, to distinguish from $Aut(E_1)$, the automorphisms of $E_1$ as an Elliptic Curve over $\mathbf{Q}$, that is, automorphisms fixing the identity point). However, $E_2$ is also a principal homogeneous space for a unique curve over $\mathbf{Q}$ with a rational point, which of course has to be $E_2$, so the cocycle $E_2$ represents could be taken to have values in $Aut(E_1)$. Now $Aut(E_1)$ is well known to be of order 6,4 or 2 depending on whether the $j$-invariant of $E_1$ is 0, 1728 or anything else, respectively. Moreover the order of the cocycle representing $E_2$ (which we now see must divide 2, 4 or 6) must be the order of the minimal field extension $K$ over which $E_1$ is isomorphic to $E_2$. So $K$ must be degree 2,3,4 or 6 unless I've made an error somewhere.
Question 2: If you restrict your focus to just elliptic curves, yes your idea is right. If it's a quadratic extension, you have exactly 1 non-isomorphic companion. If you have a higher degree number field, you have nothing but composites of the quadratic case unless your elliptic curve has j invariant 0 or 1728.
Notice I am very explicitly using your choice of the word elliptic curve for both of these answers.