Why are the Automorphisms $\text{Aut}(T_2)$ of the regular, infinite, binary rooted tree the **length-preserving** permutations? Why are the Automorphisms $\text{Aut}(T_2)$ of the regular, infinite, binary rooted tree the length-preserving permutations?
This suggests to me that only swaps of vertices (and all their children) which are at the same number of edges from the root as each other, are permissible.
But it seems to me that we could take any pair of vertices on this tree and swap them (along with all their children) and the graph would remain isomorphic to the original.
What am I misunderstanding?
 A: I'll turn my comments into an answer.
In any tree, one can assign lengths to the edges, so that all edges have length equal to $1$. Next, for any two vertices $v,w$ in the tree, there is a unique "edge-path" in the tree having endpoints $v,w$, and the length of that path is equal to the number of edges along it (just add up all the $1$'s, one for each edge in that path). 
One can prove that any automorphism $f$ of a tree preserves lengths by a simple induction argument. The basis step of the induction goes like this: if $v,w$ are connected by an edge path of length $1$, then they are connected by a single edge, so $f(v)$ and $f(w)$ are connected by a single edge, so $f(v)$ and $f(w)$ are connected by an edge path of length $1$. 
In the example of your comment, the dyadic tree with root $1/2$, we have the following relations:


*

*the vertices that are connected to $1/2$ by an edge path of length 1 (a single edge) are indeed $1/4$ and $3/4$, 

*the additional vertices that are connected to $1/4$ by an edge path of length $1$ are $1/8$ and $3/8$; 

*the additional vertices that are connected to $3/4$ by an edge path of length $1$ are $5/8$ and $7/8$;

*so, the vertices that are connected to $1/2$ by an edge path of length $2$ are $1/8$, $3/8$, $5/8$, and $7/8$.

*the edge path in this tree that connects the vertex $1/8$ to the vertex $3/4$ has length $3$:
$$\frac{1}{8} --- \frac{1}{4} --- \frac{1}{2} --- \frac{3}{4}
$$


Any automorphism of this rooted tree will fix the root $1/2$, and it will permute the two vertices $1/4$ and $3/4$, and it will permute the four vertices $1/8$, $3/8$, $5/8$, $7/8$. But those permutations are not independent of each other, because the automorphism must preserve length. So, for example, if the permutation interchanges the two vertices $1/4$ and $3/4$, and if the permutation fixes the vertex $1/8$, then that permutation is not an automorphism of the tree because it does not preserve lengths: the length of the edge-path between $1/8$ and $1/4$ and equals $1$ whereas the length of the edge-path between $1/8$ and $3/4$ equals $3$, and $1 \ne 3$.
Edited: As opposed to that negative example, here is a positive example to address your latest comment. One can define an automorphism that which does not "leave any vertex where it was" except for the root vertex. Specifically, the automorphism will take $1/4 \mapsto 3/4$, and it will take $1/8 \mapsto 5/8$, and $3/8 \mapsto 7/8$, and then one can continue defining this automorphism inductively for higher and higher powers of $2$ in the denominator. And in actuality there is a simple numerical formula for this automorphism: for any dyadic rational number $r$ we have 
$$f(r) = \begin{cases}
r & \quad\text{if $r = \frac{1}{2}$} \\
r + \frac{1}{2} &\quad\text{if $r < \frac{1}{2}$} \\
r - \frac{1}{2} &\quad\text{if $r > \frac{1}{2}$}
\end{cases}
$$
No vertex will "stay where it was", in other words no vertex (but the root) will be fixed by this automorphism.
