This question is part of Exercise 1.28 of McDuff and Salamon's Introduction to Symplectic Topology. Suppose we have a continuous function $\mu\colon\mathbb{R}\to\mathbb{R}$ which never vanishes and a constant $T$ so that $\mu(t+T)=\mu(t)$ for all $t\in\mathbb{R}$. Suppose that $\tau\colon\mathbb{R}\to\mathbb{R}$ satisfies \begin{equation} \dot{\tau}=\mu(\tau) \qquad\text{and}\qquad \tau(0)=0. \end{equation} We would then like to find a constant $T'\in\mathbb{R}$ with the property that $\tau(kT')=kT$ for all $k\in\mathbb{Z}$.
My progress: Since $\mu$ never vanishes, let's assume that $\mu >0$. Because $\mu$ is periodic, it is bounded below by some value $\alpha>0$. Then $\dot{\tau}>\alpha$, so $\tau$ is a bijection. This allows us to choose $T'\in \mathbb{R}$ so that $\tau(T')=T$, and we immediately see that \begin{equation} \dot{\tau}(T') = \mu(\tau(T'))=\mu(T)=\mu(0)=\mu(\tau(0))=\dot{\tau}(0). \end{equation} In fact we can choose $T^{(k)}\in\mathbb{R}$ for each $k\in\mathbb{Z}$ so that $\tau(T^{(k)})=kT$, and we'll have that $\dot{\tau}(T^{(k)})=\dot{\tau}(0)$. What I'm unable to do is show that $T^{(k)}=kT'$.
Some context: The periodic function $\mu$ doesn't just appear out of nowhere. We have a hypersurface $S\subset\mathbb{R}^{2n}$ which is a regular energy surface for two Hamiltonians on $\mathbb{R}^{2n}$: $S=H^{-1}(c)=H'^{-1}(c')$. Then $\lambda\colon S\to\mathbb{R}$ is defined by \begin{equation} X_{H'}(z)=\lambda(z)X_H(z) \end{equation} for each $z\in S$, where $X_H$ and $X_H'$ are the symplectic gradients of $H$ and $H'$, respectively. Finally, $z\colon\mathbb{R}\to S$ is a solution to the Hamiltonian differential equations given by $H$, and $z(t+T)=z(t)$ for each $t\in\mathbb{R}$. In the notation above, $\mu=\lambda\circ z$. The fact that $\mu$ doesn't vanish comes from the fact that $S$ is a regular surface for $H'$. I don't immediately see what other information this context might give, but I'm including it in case it gives $\mu$ some other helpful property in solving the above problem.
Consider $\tau^{-1}:\mathbb{R}\longrightarrow\mathbb{R}$. We have that:
$${(\tau^{-1})}'(s)=\frac{1}{({\tau}'\circ\tau^{-1})(s)}=\frac{1}{(\mu\circ\tau\circ\tau^{-1})(s)}=\frac{1}{\mu(s)}$$
In particular, ${({\tau}^{-1})}'$ is periodic with period $T$. This implies that in
\begin{align}\tau^{-1}(x+T)-\tau^{-1}(x)=\int_x^{x+T}{(\tau^{-1})}'(s)\,ds&&&&\text{(FTC)}\end{align}
the RHS does not depend on $x$. Moreover, notice that $\tau^{-1}(0)=0$. Therefore, letting $T'=\tau^{-1}(T)=\int_0^{T}{(\tau^{-1})}'(s)\,ds$, we have that $\forall x \in \mathbb{R}$
$$\tau^{-1}(x+T)-\tau^{-1}(x)=T'$$
In particular, $\tau^{-1}(kT)=kT'$, or $\tau(kT')=kT$ as desired.