Alternate definition of fields

Question

Is it true that we can define a field as a group $G$ that is isomorphic to its automorphism group?

I believe it's an equivalent definition if addition is the group operation of $G$ and multiplication is the application of the automorphism corresponding to the left element to the right element.

Answer 1

No, this is wrong for a whole bunch of reasons. If $F$ is a field, there is a map from the set of nonzero elements of $F$ to the group of automorphisms of the additive group $(F,+)$. However, this does not mean that $(F,+)$ is isomorphic to its own automorphism group, for quite a lot of different reasons:

1. $0$ is not included (multiplication by $0$ is not an automorphism of $(F,+)$).
2. Even ignoring $0$, the map is not a homomorphism from $(F,+)$ to its automorphism group. Adding two elements of $F$ does not correspond to composing the automorphisms given by multiplication by them. Rather, the map is a homomorphism from the group $(F\setminus\{0\},\cdot)$ (with multiplication as the group operation) to the group of automorphisms of $(F,+)$.
3. The map in question is typically not surjective. That is, $(F,+)$ may have lots of other automorphisms besides those that are given by multiplication by elements of $F$.

(And this is not even getting into the issues you would have going "the other way", starting with a group isomorphic to its automorphism group and getting a field. For instance, such a group need not be abelian.)