Automorphism Group of Bruhat-Tits Tree Let $K$ be a discretely valued complete field extension of $\mathbb{Q}_p$.
In Casselman's notes on the Bruhat-Tits tree $T$, exercise 4.3 claims that all automorphisms of $T$ are induced by elements of $\text{GL}_2(K)$.
However, in Serre's "Trees", exercise 1 on page 78 claims that the homomorphism
$$
\text{GL}_2(K) \rightarrow \text{Aut}(T),
$$
is not surjective. This is also implied by this answer.
Which is correct? If the latter, is there a simple example of an automorphism of $T$ not induced by a matrix?
(strictly Casselman's claim is that all isometries are induced by matrices, but it seems all automorphisms are isometries)
 A: To give an example of an automorphism that is not induced by a matrix, note that the points of distance 1 away from the central point can be identified with $ \mathbb{P}^1( \mathfrak{o}/\mathfrak{p})$, and any matrix which stabilises the central point is an element of $ \mathrm{GL}_2(\mathfrak{o})$. It is an fun exercise to show that the induced action on the projective space is in fact the action by Mobius transformations
$$
\mathrm{GL}_2(\mathfrak{o}) \rightarrow \mathrm{GL}_2(\mathfrak{o}/\mathfrak{p})\rightarrow \mathrm{Aut}(\mathbb{P}^1(\mathfrak{o}/\mathfrak{p}))
$$
and this final map is not a surjection (as in the case over $\mathbb{C}$, such an automorphism is specified by the action on 3 points). However there are clearly automorphisms of $T$ inducing any set theoretic automorphism of $\mathbb{P}^1(\mathfrak{o}/\mathfrak{p})$. Note that this argument doesn't work when $q=2$ or $q=3$, you will have to go to distance 2 away!
Therefore a concrete example of an automorphism of the tree for $p=5$ not arising from matrix multiplication would be to 'swap' two branches and leave the other three branches as they are.
