Let $E \subset \mathbb{P}_\mathbb{C}^2$ be an elliptic curve. If $E$ has complex multiplication (by anything) then the theory of complex multiplication in particular tells us that if $\sigma \in \textrm{Aut}(\mathbb{C})$ then $E^\sigma$ will be isogenous to $E$. Suppose $\textrm{End}(E) \cong \mathbb{Z}$ do we still have that $E^\sigma \cong E$ or are there some obvious counterexamples? What if we insist that the $j$-invariant, $j(E)$ is an algebraic number?
[Math] Isogeny classes of elliptic curves
arithmetic-geometryelliptic-curvesnt.number-theory
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I don't know anything about hypergeometric functions, so this is not a direct answer to your question. But, I have thought a lot about the problem of detecting complex multiplication of elliptic curves (and certain higher-dimensional analogues for abelian varieties).
Suppose you are given an algebraic integer $j$, and you wish to know whether it is a CM j-invariant. Then there is a sort of night-and-day algorithm you can perform here, where by night you reduce the elliptic curve modulo various primes and keep in mind the fact that if your elliptic curve has CM, then half the time (in the sense of density) you will get a supersingular elliptic curve and the other half you will get a CM elliptic curve whose characteristic p endomorphism algebra is the same as the algebra you started with. Thus, in practice, if your curve does not have CM, you will fairly quickly be able to rule it out by finding two primes of ordinary reduction with different endomorphism algebras. So far this is just a probabilistic algorithm. Once you figure out that the CM field is either a particular quadratic field K or there is no CM at all, you compute (e.g. by classical CM theory as described e.g. in Cox's book Primes of the form...) you compute the $j$-invariants of elliptic curves with K-CM. There are infinitely many of these, because the j-invariant depends on the endomorphism ring (equivalently, the conductor of the order), but you can either just compute all of them in order of conductor or look more carefully at the mod p reductions and get a bound on what the conductor could be.
[Edit: Actually, you can figure out exactly what the CM order must be by computing the endomorphism ring at any two primes of ordinary reduction. It is a theorem that if $E$ is a curve with CM by the order of conductor $f$ in a CM field $K$ and $p$ is a prime of ordinary reduction, the conductor of the reduced endomorphism ring is $f/p^{ord_p(f)}$, i.e., you just strip away the $p$-part of the conductor.]
This is not the state of the art, though. Rather, see the paper
Achter, Jeffrey D. Detecting complex multiplication. (English summary) Computational aspects of algebraic curves, 38--50, Lecture Notes Ser. Comput., 13, World Sci. Publ., Hackensack, NJ, 2005.
[A copy is available via his webpage http://www.math.colostate.edu/~achter/.]
In the paper, Achter uses Faltings' theorem and the effective Cebotarev density theorem to eliminate the "day" part of the algorithm. He also gives a complexity analysis and explains why this is faster than what I sketched above.
Finally, I'm sure the questioner knows this, but others may not: for elliptic curves over $\mathbb{Q}$ there's no need to do any of this. Rather you just compute the $j$-invariant and see whether it's one of the $13$ $j$-invariants of CM elliptic curves over $\mathbb{Q}$ associated to the $13$ class number one quadratic orders (yes, this relies on the Heegner-Baker-Stark resolution of Gauss' class number one problem). For the list, see e.g.
modular.fas.harvard.edu/Tables/cmj.html
Best Answer
We say that an elliptic curve $E$ over a number field $K$ is an elliptic $\mathbf{Q}$-curve if it is is isogenous to its Galois conjugates $E^\sigma$. These were first studied by Benedict Gross, but were later studied by Elkies, Ellenberg, Ribet, and many others. It's possible to show that $\mathbf{Q}$ curves are modular, independently of the BCDTW modularity theorem, and that even though they are defined over $K$, their $p$-torsion Galois representations (at least away from the degrees of said isogenies) descend down to $\mathbf{Q}$. Elkies also famously showed if $E_{/K}$ is a $\mathbf{Q}$-curve without CM, there must be a geometrically isogenous curve $E'_{/L}$ such that $Gal(L/\mathbf{Q}) \cong (\mathbf{Z}/2\mathbf{Z})^r$ for some integer $r$. This comes down to showing that the moduli space of elliptic $\mathbf{Q}$-curves of degree $N$ form a certain very special quotient of $X_0(N)$. For details on most of this, see Jordan Ellenberg's "$\mathbf{Q}$-curves and Galois representations" ( http://www.math.wisc.edu/~ellenber/MCAV.pdf )