I have something that is driving me crazy for some while. In Katok and Hasselblat's book, Introduction to Modern Theory of Dynamical Systems, on page 65 they go through a discussion about time change. On page 65 they write an equation (2.2.5). I am able to drive very similar relations between $\beta$ and $\alpha$ but never the less, no matter what I do, I can't drive the exact same relation they get in (2.2.6). Could someone please tell me how they do get to relation (2.2.5)? I have a feeling that they messed up something but I'm not quite sure.
I now go through what I do. By assumption, we are doing a time change with the reparametrization $\alpha$ to get a flow $\psi^t := \varphi^{\alpha(t,\ .)}$. Also by assymption, $\psi^t$ and $\varphi^t$ are conjugate by the same $h$ for all $t \in \mathbb{R}$: $$ h\varphi^th^{-1} = \psi^t; \quad \forall t \in \mathbb{R}. $$
Hence for any $x$ $$ h(\varphi^t(x)) = \psi^t(h(x)). $$ But \begin{align} \psi^t(h(x)) &= \varphi^{\alpha(t,\ h(x))}h(x) \\ &= \varphi^{\alpha(t,\ h(x)) + \beta(x)}(x) \end{align} \begin{align} h(\varphi^t(x)) &= \varphi^{\beta(\varphi^t(x))}(\varphi^t(x)) \\ &= \varphi^{\beta(\varphi^t(x)) + t}(x). \end{align}
Therefore $\beta(\varphi^t(x)) + t = \beta(x) + \alpha(t, h(x))$, i.e. $$ \alpha(t, h(x)) - t = \beta(\varphi^t(x)) - \beta(x). $$
I really suspect that $\alpha(t, h(x)) = \alpha(t, x)$ so I think what they wrote is wrong but I might be mistaken. Anyways, differentiating with respect to $t$ gives:
$$ a(h(x)) - 1 = \xi(x)(\beta); \quad \forall x $$ or to make things shorter, $$a \circ h - 1 = \xi(\beta).$$ This is also different from what they have: $$ a - 1 = \xi(\beta). $$
However (I'm not sure) that I once proved $a$ and $\beta$ are invariant under iterations of $h$ which means that in that case the equation they have is also correct. But nevertheless i don't really see how they get directly into it.