Relationship between the sizes of orbits of nodes

About a year and a half ago I wrote a certain proposition offhand in a proof. I must have thought it was trivial at the time and so did not prove it, but months later it is not quite so trivial to me anymore.

Proposition. Let $T$ be a finite free tree (a connected acyclic graph). If $u$ and $v$ are adjacent in the tree, then the size of the orbit of $u$ is an integer multiple of the size of the orbit of $v$ under the group of all graph automorphisms of $T$, or the other way around.

I hope this statement is actually true, as I have some further results that rely on it. Would anyone be able to help me with a proof?

Solution 1:

Let $u$ be a leaf and $v$ it's unique neighbour. Let $G = Aut(T)$. By orbit-stabilizer theorem: $$|G| = |G\cdot u| \hspace{4pt} |G_u| = |G\cdot v| \hspace{4pt}|G_u|$$ Every stabilizer of $u$ must permute it's neighbours but since $u$ has a single neighbour, $G_u \subseteq G_v$. By Lagrange's Theorem, $|G_u|$ divides $|G_v|$, so we have our result by the orbit-stabilizer.

Now for arbitrary neighbours $u$ and $v$, we can iteratively "snip the leaves" until we obtain a subtree $T'\subseteq T$ with at least one of $u$ or $v$ a leaf. It's clear that $G':= Aut(T') \subseteq Aut(T)$, $G'_u \subseteq G_u$, and $G'_v \subseteq G_v$, so a final application of the orbit-stabilizer and Lagrange's theorem does the trick.