How do I calculate the automorphism of a tree with 7 vertices?

Question

I found this question which explains how it is done for 6 vertices (https://math.stackexchange.com/a/420259/515460), however I do not understand how $|Aut(G)|$ has been calculated?

Answer 1

As mentioned in the comments, the example you're interested in (where the tree is a path) is straightforward, and $|\text{Aut}(G)| = 2$.

In general, we can solve the problem by thinking about matching up identical-looking leaves. Once leaves are fixed, nothing else needs to be done.

I will steal the six examples of $6$-vertex trees from the question you linked to. Yoink!

In each example, let's number the vertices $1, 2, 3, \dots, 6$ starting from top right clockwise, to make talking about specific vertices easy.

1. In the first tree, all five leaves $\{1,3,4,5,6\}$ are identical, and we can permute them any which way. There are $5!$ automorphisms.

2. In the second tree, leaf $4$ is doing its own thing, and we can permute leaves $\{1,5,6\}$ however we like. There are $3!$ automorphisms.

3. In the third tree, leaves $1$ and $6$ are interchangeable, but distinct from leave $5$, so there are $2$ automorphisms.

4. In the fourth tree, we can map leaf $1$ to any of the leaves $\{1,4,5,6\}$, but then $4$ must be mapped to the leaf closest to $1$, and there will be two choices for how to arrange $5$ and $6$. $4 \cdot 2 = 8$ automorphisms. Or to put it another way: the leaves come in two bunches of two leaves, and we can swap the bunches and permute the leaves in each bunch.

5. In the fifth tree, leaves $5$ and $6$ are interchangeable and different from $1$. There are $2$ automorphisms.

6. In the last tree, the only two leaves $1$ and $6$ are interchangeable. There are $2$ automorphisms.

This basically covers all the complexity you expect to see in a tree's automorphisms, though sometimes several of these things will happen independently.