Automorphisms of the rooted tree operad

Question

This follows Ryan Budney's comment to the question asked here.

What is the automorphism group of the rooted tree operad?

(By the rooted tree operad, I just mean the operad with object rooted trees and morphisms given by grafting a root to a leaf).

Answer 1

I think the answer to the question as literally stated is "the trivial group", but I think there are related inquiries which get into some deep combinatorics.

One way of thinking about the rooted tree operad is that it is the free operad $O(F)$ generated by the Joyal species $F$ (a functor $\mathbb{P} \to Set$ where $\mathbb{P}$ is the groupoid of finite sets $\{1, \ldots, n\}$ and permutations) where $F(0)$ is empty and $F(n)$ is a singleton for $n \geq 1$. You can think of the element of $F(n)$ as a "sprout" $s_n$ consisting of a root, $n$ leaves, and no other nodes, and then the elements of $O(F)$ are obtained recursively by starting with sprouts and applying grafting operations.

So we're looking at operad automorphisms $\phi: O(F) \to O(F)$. By freeness, the endomorphisms of $O(F)$ are in bijection with natural transformations $\psi: F \to U O(F)$ where $U O(F)$ is the underlying species or permutation representation of $O(F)$. Concretely, to give such a natural transformation is to give a collection of trees $t_n = \psi_n(s_n)$ for all $n \geq 1$ where each $t_n$ must be invariant under permuting the leaves, since the sprout $s_n$ is invariant under such permutations. That's a pretty strong condition on $t_n$, and there are actually precious few such collections.

But now you want more: you want $\phi$ to be an automorphism as well. So each sprout $s_n$ must be in the image of $\phi_n$. But no nonsprout tree $u$ can ever map to $s_n$ under $\phi_n$, because if $u$ is obtained by grafting together more than one sprout $s_k$, then $\phi_n(u)$ is obtained by similarly grafting together more than one tree $t_k$, and this is never a sprout.

So in order for there to exist $u$ such that $\phi_n(u) = s_n$, we must have $u = s_n$. To have that for all $n$ means $\psi(s_n) = s_n$ for all $n$, hence the only operad automorphism is the identity automorphism.

I think a more interesting inquiry is to understand the groupoid of rooted trees and isomorphisms between them. This is an incredibly rich object.

Edit: Let me make my last suggestion more precise. Let's define a rooted tree to be a finite set $X$ equipped with a function $f: X \to X$ and an element $r \in X$ such that $f^{(n)}(X) = \{r\}$ for sufficiently large $n$. The idea is that $f(x)$ is one step closer to the root than $x$, unless $x$ is the root. Then an isomorphism is a function $\phi: X \to Y$ which preserves the stepping-closer function and the root. It is determined by its restriction to the leaf set.

But even this groupoid isn't that mysterious; it seems automorphism groups are iterated wreath products of symmetric groups. Here is a different but related inquiry which I think is rather more interesting: regarding trees $T$ as posets, describe the category of order-preserving bijections between trees.