Let $E/K$ be an elliptic curve over a field $K$. Let $Aut(E)$ be a group of all automorphism of $E$ as algebraic curves. Then, does $AutE\cong E$ holds ?
There is a natural injection $E\to \text{Aut}(E)$ given by $P\mapsto \{E\to E, Q\mapsto Q+P\}$.
To prove this is surjective, we should prove every automorpshims of $E$ is given by translation.
Background question:
Let $\text{Aut}(E,+)$ be an set of all isomorphism which commutes with $E$ action to $E$ itself (as trivial torsor).
Then, it is known that $\text{Aut}(E,+)\cong E$
(cf. lem$1.3$ of https://arxiv.org/abs/math/0606580).
But I wonder why condition of 'commute with action' needed to prove this isomorphism, so I asked this question.
The isomorphism $AutE\cong E$ is never true. First of all, any elliptic curve also has an inversion map, sending $P$ to $-P$, which is not given by any translation. There also exist elliptic curves with complex multiplication, which have additional automorphisms besides these, though these are only finitely many (at most $24$ in characteristic $2$, $12$ in characteristic $3$ and at most $6$ otherwise). See Silverman, section III.10 for a discussion of these additional automorphisms.
Any automorphism of an elliptic curve fixing the origin is an isogeny, see Silverman section III.4. Therefore the automorphism group of $E$ as an algebraic curve will be the semidirect product of $E$ and the automorphisms of $E$ as an elliptic curve.