Barycentric subdivision removes edge inversions while preserving the tree

Every automorphism of $T$ permutes geometric edges, so it sends the midpoint of an edge to the midpoint of its image edge. Hence the original action extends uniquely to the subdivided graph $T^{\mathrm{\prime }}$.

The graph $T^{\mathrm{\prime }}$ is connected because each original edge has only been split in two. A reduced closed path in $T^{\mathrm{\prime }}$ would project, after deleting midpoint backtracks, to a nontrivial reduced closed path in $T$, contradicting [L2]. So $T^{\mathrm{\prime }}$ is again a simplicial tree.

In $T^{\mathrm{\prime }}$, every half-edge joins an original vertex to a midpoint vertex. Its reverse joins a midpoint vertex to an original vertex, so an automorphism cannot send a half-edge to its reverse without swapping the two vertex types. That cannot happen in the extended action. Therefore the action on $T^{\mathrm{\prime }}$ is without inversions in the sense of [L1].