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>



Automorphism of the transfinite rooted binary tree
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.


