To better illustrate what I'm trying to say, here's an example. Path $\gamma_1(t)=(cost,-sint)$, $t\in[0,2\pi]$ is a (negatively oriented) parametrization of circle, with starting and ending point $(1,0)$, and $\gamma_2(t)=(-cost,sint), t\in[0,2\pi]$ is also (negatively oriented) parametrization of the same circle, but with starting and ending point $(-1,0)$. Traces of these two parametrizations are the same, and they also have the same orientation. But their starting and ending points are not the same, that is these paths are not "connecting the same two points". Are these two smooth paths equivalent?
(I'm looking for an answer in general, not specifically for the example I have given.)
(We defined parametrized curve as a map $\gamma:I\rightarrow\Bbb R^n$ of class $C^1$, $I\in\Bbb R$ any interval. If $I=[a,b]$ (that is, $I$ is closed segment), then we also call it a smooth path. Two parametrized curves $\alpha:I\rightarrow\Bbb R^n$ and $\beta:J\rightarrow\Bbb R^n$ are equivalent if there exists $C^1$-diffeomorphism $\phi:I\rightarrow J$ such that $\beta=\alpha \circ\phi$ (that is, $\beta$ is reparametrization of $\alpha$). Trace is the image of a parametrized curve $\gamma$, so we also say that $\gamma$ is parametrization of its trace. Apologies for any bad translations!)