Automorphism group of the class of surreal numbers Do we know the group (Group) of automorphisms of the ordered Field of surreal numbers? 
I feel the different ways to see the surreal numbers should provide us with several ways to define interesting automorphisms. However, I can't think of any except the identity.
 A: To avoid having to worry about things like proper classes, let us suppose $\kappa$ is an inaccessible cardinal and that by "the surreal numbers" $No$ we mean the surreal numbers of rank $<\kappa$.  Then the automorphism group of $No$ is truly enormous: it has cardinality $2^\kappa$, and is what Conway calls an "improper class".  This essentially follows from Theorems 28 and 29 of ONAG.  Specifically, if $f\subseteq g\subset No$ and $\bar{f}\subset No$ are small subfields (i.e. of cardinality $<\kappa$) and $\varphi:f\to \bar{f}$ is an isomorphism of ordered fields, then Theorem 28 says there is a subfield $\bar{g}$ such that $\bar{f}\subseteq \bar{g}\subset On$ and an isomorphism $\varphi':g\to \bar{g}$ extending $\varphi$.  Furthermore, note that for any such $f$ and $\varphi$, there is an $g$ such that the extension $\varphi'$ is not unique (for instance, if $g=f(x)$ where $x$ is greater than every element of $f$, then $\varphi'$ can send $x$ to any $x\in No$ which is greater than every element of $\bar{f}$).  As in the proof of Theorem 29, we can construct automorphisms of $No$ by iterating this extension property back and forth by an induction of length $\kappa$, and we can do so such that at $\kappa$ many steps of the iteration, we have two different ways to choose the extension.  Thus for every function $\kappa\to\{0,1\}$ (telling us how to make the choices), we get a different automorphism of $No$.
A: This question is explored in detail in "All Numbers Great and Small," which can be found in Real Numbers, Generalizations of the Reals, and Theories of Continua (pp. 239-258).
Regarding automorphisms of $\mathsf{No}$ considered as an ordered field, there is, in fact, a proper class of such. Or, put in a (perhaps) more palatable way, given any set of ordered field automorphisms of $\mathsf{No},$ there is an ordered field automorphism of $\mathsf{No}$ not contained in the given set.
However, the identity map is the only ordered field automorphism that respects the tree structure inherent in the construction of $\mathsf{No}.$ That is, if we define the relation $<_S$ by $x<_S y$ (read "$x$ is simpler than $y$") iff $x\in L_y\cup R_y$ (where $L_y,R_y$ are the respective sets of left and right options of $y$), then the identity map is the only automorphism on $\mathsf{No}$ that is $<_S$-preserving (this is the first theorem of $\S5$).
