Define a parametrized curve as $r:(a,b)\to\Bbb R^3$ be $C^\infty$. It is clear that a curve can have different parametrizations. Say, $(u-v)t+u$ and $(v-u)t+v:t\in(0,1)$ parametrized the same line segment connecting $u,v$.
I am learning elementary differential geometry. I got a definition of reparametrization of a curve as a composition of a curve and a deffeomorphism. i.e if $r:(a,b)\to\Bbb R^3$ is a curve, a reparametrization is a $r \circ \phi:(c,d)\to \Bbb R^3$, where $\phi:(c,d)\to (a,b)$ is a deffeomorphism.
I am wondering if $r,l$ are parametrizations of a curve $C$, is it necessary for $r$ to be a reprarametrizaiton of $l$? for example, $(u-v)t+u$= $(v-u)\phi(t)+v:t\in(0,1)$, where $\phi(t)=1-t$. But I have trouble finding $\phi$ for general curves.
There is a counterexample given by two parametrizations of the first factor $\mathbb{R}$ in $\mathbb{R}^3$, namely
$$r:x\mapsto (x^3,0,0)\text{ and }l:x\mapsto (x,0,0).$$
If there is a diffeomorphism $\varphi$ such that $l=r\circ\varphi$, i.e. $\forall x\in\mathbb{R}, x=\varphi(x)^3$, then $\varphi(\bullet)=\sqrt[3]{\bullet}$. But this latter is not a diffeomorphism, since it is not differentiable at $x=0$.