Automorphism of the transfinite rooted binary tree I was studying combinatorical group theory recently, and I came across the infinite regular rooted binary tree and its automorphism group $Aut(T^{(2)})$with the Grigorchuk subgroup.
Let me now elaborate more on a different topic, the surreal numbers. They are constructed similarly to the binary tree at first, but in the $\omega$-th generation, infinitesimal and transfinite numbers start to appear.
I wonder - if we remove, from the surreal number "tree", all numbers and order relations, thus focusing solely on the "transfinite binary tree" (in a graph-theoretic sense), what would be the automorphism group (I allow proper classes to be groups) of that tree?
For a better vsiualization of what I mean by such a tree:
https://upload.wikimedia.org/wikipedia/commons/thumb/4/49/Surreal_number_tree.svg/532px-Surreal_number_tree.svg.png
 A: Although this question already has an accepted answer, which is correct for the question as stated, I posit that the surreal number tree is best viewed as a tree in the order-theoretic rather than the graph-theoretic sense.
In that case, the automorphism group is isomorphic to the direct product of a class of two-element groups indexed by the surreal numbers:
$$ G := \prod_{\alpha \in \textbf{No}} C_2 $$
To formalise this, consider an automorphism $\phi : \textbf{No} \rightarrow \textbf{No}$. We describe the surreal numbers inductively, where element is either:


*

*The zero (root) number $0$;

*The left successor (child) $l(\alpha)$ of an existing number $\alpha \in \textbf{No}$;

*The right successor (child) $r(\alpha)$ of an existing number $\alpha \in \textbf{No}$;

*A limit of an ascending chain (with respect to the partial order induced by the tree, not numerical order) of numbers $\sup_{i \in I} \alpha_i$.


We shall construct such a $\phi$ naturally from an arbitrary $\psi \in G$ viewed for convenience as a function $\psi : \textbf{No} \rightarrow C_2$. Let $C_2$ act on the two-element set $\{ l, r \}$ in the obvious way, and define:


*

*$ \phi(0) := 0 $

*$ \phi(l(\alpha)) := (\psi(\alpha)(l))(\phi(\alpha)) $

*$ \phi(r(\alpha)) := (\psi(\alpha)(r))(\phi(\alpha)) $

*$ \phi(\sup_{i \in I} \alpha_i) := \sup_{i \in I} \phi(\alpha_i) $


We have described a map $\Theta : G \rightarrow \textrm{Aut}(\textbf{No})$ which sends $\psi$ to $\phi$, and it is straightforward to check that this is an isomorphism.
EDIT: I should probably clarify that in order to manipulate the surreals set-theoretically without pain, we can think of them as being a subset of a Grothendieck universe within our ambient set theory. It is henceforth safe to treat the surreals of that Grothendieck universe as being a set within the larger ambient set theory, and perform usual set-theoretic constructions such as powersets.
A: Each connected component of the tree is either rooted at $0$ or at a number which birthday is a limit ordinal, these are exactly the vertices of degree $2$. Let $A$ be the class of all limit ordinals, with addition of $0$. For an element $\alpha \in A$ let $s_\alpha =  \{+, -\}^\alpha$ be the set of its sign expansions, and $\mathcal{A} = \cup_{\alpha \in A} s_{\alpha}$ the class of "roots". In a way, $\mathcal{A}$ is a quotient of the class of surreal numbers $\mathbf{On}$ and $\mathfrak{c} = 2^{\aleph_0}$ (the size of a component).
The automorphism group is now the semidirect product of $G^{\mathcal{A}}$ and $\mathrm{Sym}(\mathcal{A})$, where $G$ is the group of automorpisms of the infinite tree of depth $\omega$. Of course, you need to adopt a powerful enough set theory for any of this to make sense.
