 Solution

The assertion is false because excursion coding does not remember the speed of traversal. Let $f$ be any nonzero coding function and let $\varphi :[0,1]\to [0,1]$ be a nonidentity increasing homeomorphism. Set $g=f\circ \varphi$. ThenThus $[s]\mapsto [\varphi (s)]$ induces an isometry $T_g\to T_f$, although generally $g\ne f$.