Let $(X_t, t \in \mathbb{R})$ be a real Markov process, $R \subset \mathbb{R}_+$. $h_t$ is Borel-measurable bijection and its inverse is also Borel-measurable. Then $Y_t = (h_t(X_t), t)$ is also a Markov process.
My attempt: Consider one of the equal definitions of a Markov process: $\forall m, \forall s_1 < ... < s_m < s < t, \forall f$ - bounded Borel function $ E(f(X_t)|X_s, X_{s_m}, ..., X_{s_1}) = E(f(X_t)|X_s)$. Then let's take any $f$ and $f(h(X_t))$. Composition of Borel functions is a Borel function, so the process is Markov.
But where to use the fact that h is a bijection?