Finite groups acting on trees have a global fixed vertex after subdivision

Replace $T$ by its barycentric subdivision $T^{\mathrm{\prime }}$ using [L1]. Choose a vertex $v$ of $T^{\mathrm{\prime }}$. Because $G$ is finite, the orbit $G\cdot v$ is finite, and the union of the geodesics joining pairs of orbit vertices is therefore a finite $G$-invariant subtree $U\subseteq T^{\mathrm{\prime }}$.

Let $D$ be the diameter of $U$, choose vertices $x,y\in U$ with $d(x,y)=D$, and let $m$ be the midpoint of the unique geodesic from $x$ to $y$. In a finite tree every diameter geodesic has the same midpoint: if two diameters had different midpoints, the unique path joining those midpoints would extend one of them past length $D$, contradicting maximality. So $m$ depends only on $U$, not on the chosen diameter.

Any automorphism of $U$ sends diameter geodesics to diameter geodesics, so it fixes the intrinsic midpoint $m$ from step 2.1. If $D$ is even, $m$ is a vertex of $U$ and is fixed. If $D$ is odd, $m$ is the midpoint of a unique geometric edge of $U$, so every element of $G$ preserves that edge setwise. The action on $T^{\mathrm{\prime }}$ is without inversions by [L1], hence no element can swap the two endpoints; both endpoints are fixed. In either case the induced action on $T^{\mathrm{\prime }}$ fixes a vertex.