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.
Theorem (originally due to Setzer?): Fix $E/\mathbb{Q}$ with $j(E)$ not 0 or 1728. Then for all but finitely many inequivalent twists $E_d$, the torsion subgroup $E_d(\mathbb{Q})_{tors}$ is isomorphic to $E[2](\mathbb{Q})$, so in particular $E_d(\mathbb{Q})_{tors}$ has order 1, 2, or 4. (Probably he also proved it for number fields.)
There's a paper of mine$^1$ with a much more general theorem using the theory of heights. I don't recall Setzer's proof except that it doesn't use heights.
Theorem: Let $K$ be a number field and let $A/K$ be an abelian variety with $\mu_n\subset {\rm Aut}(A)$. (This means we can twist $A$ by $n$'th roots of $d$.) Then every point $P\in A_d(K)$ satisfies one of the following two conditions:
- $P$ is fixed by a non-trivial $\zeta\in\mu_n$.
- $\hat h(P) \ge C_1(A)h^{(n)}(d) - C_2(A)$.
Here $\hat h$ is the canonical height relative to an ample symmetric divisor, and $h^{(n)}(d)$ is a sort of "$n$'th power free height," say equal to the minimum of $h(du^n)$ for $u\in K^*$. The constants depend on $A$, but are independent of $d$.
It follows from the theorem that after discarding finitely many $d \in K^*/{K^*}^n$, a point in $A_d(K)$ is either $1-\zeta$ torsion (hence $nP=O$), or its canonical height is positive, and hence it is nontorsion.
Of course, to describe more precisely what happens for the finitely many exceptional $d$ can be a delicate matter, as some of the other answers have indicated. I think it's interesting to see how one can approach the problem via heights or via representation theory.
$^1$ J.H. Silverman, Lower bounds for height functions, Duke Math. J. 51 (1984), 395-403.
EDIT: Fixed statement of first theorem. I'd originally written that "for all but finitely many inequivalent twists $E_d$, the torsion subgroup $E_d(\mathbb{Q})_{tors}$ has at most two elements." This is clearly false, since if $E$ has the form $E:y^2=(x-a)(x-b)(x-c)$ with $a,b,c\in\mathbb{Q}$, then $E_{d}[2](\mathbb{Q})$ has order 4 for every twist.
Best Answer
These are exactly the primes where the Neron model of $E$ has fiber type, in Kodaira's table, anything other than $I_0$ and $I_0^*$. Here $I_0$ represents good reduction and $I_0^*$ represents a quadratic twist of good reduction by a ramified extension (quadratic twisting by an unramified extension preserves the reduction type).
(At least, this is true away from primes above 2, which are surely more complicated).
The fiber type can be computed by Tate's algorithm, but to tell whether the type is $I_0$ or $I_0^*$, you can simply perform a ramified quadratic twist and check whether either the curve or its quadratic twist has good reduction. The number you quadratic twist by doesn't matter, as long as it is $\pi$-adic valuation, where $\pi$ is the prime of bad reduction, is odd.
For $\pi$ in $\mathbb Q(\sqrt{-5})$ lying over a prime $p$ of $\mathbb Q$, unless $\pi$ is ramified (i.e. $\pi$ lies above $5$), then any number which has $p$-adic valuation odd will do the trick.