Can a change of variables in a dynamical system result in a solution that can not be mapped back to the original solution? Specifically, consider the dynamical system
\begin{align} x^\prime(t) &= f(x(t)), &x(0) &= x_0 \end{align}
where $x(t)\in\mathbb{R}^m$. Given an invertible map $\phi : \mathbb{R}^m\rightarrow\mathbb{R}^m$ where $y(t)=\phi(x(t))$, we can define a new system
\begin{align} y^\prime(t) &= \phi^\prime(x(t))x^\prime(t) \\ &= \phi^\prime(x(t))f(x(t)) \\ &= \phi^\prime(\phi^{-1}(y(t)))f(\phi^{-1}(y(t)) \end{align}
or that
\begin{align} y^\prime(t) &= \phi^\prime(\phi^{-1}(y(t)))f(\phi^{-1}(y(t))), &y(0) &= \phi(x_0) \end{align}
Does the system for $x$ have a solution if and only if $y$ has a solution? If so, can we always map one solution to the other using $\phi$?
Yes. Use the fact that $\phi$ is invertible and apply the rule of differentiation of the inverse function.
For simplicity I only consider the case $\phi:\mathbb R\to \mathbb R\,$:
Let $x$ be a solution of $\dot x=f(x)\,.$ Then $y=\phi(x)$ is a solution of $$ \dot y=\phi'(x)\,\dot x=\phi'(\phi^{-1}(y))\,f(\phi^{-1}(y))\,. $$ Conversely, let $y$ be a solution of that last system which we write more briefly as $$ \dot y=g(y) $$ where $g(y)=((\phi'f)\circ\psi)(y)$ and $\psi:=\phi^{-1}\,.$ Then $x=\psi(y)$ is a solution of $$ \dot x=\psi'(y)\,\dot y=\psi'(\phi(x))\,g(\phi(x))=\frac{1}{\phi'(\psi(\phi(x)))}((\phi'f)\circ\psi)(\phi(x))=\frac{\phi'(x)f(x)}{\phi'(x)}=f(x)\,. $$