What's the automorphism group of the real and complex numbers and the quaternions? According to Wikipedia the automorphism group of the octonions is the exceptional group $G_2$. Are there analogous groups for the real numbers, the complex numbers and the quaternions?
 A: Regarding $R,$ let $h:F \to  G$ be an isomorphism between subfields $F,G$ of $R .$
If  $\forall x\in F\; (x>0\implies \sqrt x\in F),$ then $h=id_F. $ Because, for $x,y\in F,$ we have $$x>y\implies 0\ne h(x)-h(y)=(h(\sqrt {x-y}))^2 \implies h(x)>h(y).$$ And $h|Q=id_Q.$ So $\{q\in Q:q<x\}=\{q\in Q:q<h(x)\}$ for all $x\in F.$ In particular if $F=R$ then  $h=id_R$  and $G=R.$
A: The automorphism group of $\mathbb{R}$ is trivial.
The automorphism group of $\mathbb{C}$ is extremely complicated (at least if you accept the axiom of choice) : see http://www.jstor.org/stable/2689301?seq=1#page_scan_tab_contents for instance.
The automorphism group of $\mathbb{H}$ is $\mathbb{H}^*/\mathbb{R}^*$ : according to the Skolem-Noether theorem, every $\mathbb{R}$-algebra automorphism of $\mathbb{H}$ is inner, so it just remains to quotient out the center. But every ring automorphism must induce an automorphism of the center, and since the center $\mathbb{R}$ does not have not-trivial automorphisms, every ring automorphism is actually a $\mathbb{R}$-algebra automorphism.
Note that this group is actually isomorphic to $SO(3)$ : the conjugation action is by isometry on the subspace of pure quaternions.
