$\def\RR{\mathbb{R}}
\def\vr{\mathbf{r}}
\def\vv{\mathbf{v}}
\def\vnull{\mathbf{0}}
\def\ss{\subseteq}
\def\To{\rightarrow}
\def\p{\pi}$Consider a curve in $\RR^n$ given by
$\vr(t) = (x_1(t),\ldots,x_n(t))$ for $t\in I\ss\RR$.
Suppose $\vr$ is differentiable and non-self intersecting.
(The restriction of non-self intersecting can be relaxed, but introduces technicalities that we wish to avoid.)
Let $\vv(t) = \vr'(t)$ and further suppose that $\vv(t)\ne \vnull$ for any $t$.
The arclength along this curve will be given by
\begin{equation}s=\int_{t_0}^t |\vv(t)|dt,\tag{1}\end{equation}
where $t_0\in I$ is some reference value for the parameter $t$ corresponding to an arclength of $s=0$.
(Note that $t<t_0$ corresponds to a negative arclength.
This sign agrees with the orientation of the curve.)
By assumption, $|\vv|>0$.
Thus, $s$ is a strictly monotone increasing function of $t$ and so will have an inverse function, $t=t(s)$.
Thus,
$\vr(s)\equiv\vr(t(s))$,
where $s\in I'\ss\RR$,
will be this same curve parametrized by arclength.
The interval $I'$ can be determined from (1).
After this procedure has been done, it should be clear that two parametrizations of the same curve,
$\vr_1(s)$ and $\vr_2(s)$,
can differ in their representation only through the following transformation:
$s\To ks+s_0$,
where $k=\pm 1$.
That is, the curves can only differ by a shift in $s$ or by having opposite orientations.
(It is also possible that after this transformation further transformations on $I'$ may be allowed and necessary, depending on the periodicity of the components of $\vr(s)$.)
A quick way that one can distinguish different curves is by examining the intervals $I'$.
For example, if the intervals are finite subsets of $\RR$ and are different in length, then the curves are not the same. (They do not have the same length.)
Examples:
Let
\begin{align*}
\vr_1(t) &= (\sin t,\cos t),
& t\in[0,2\p) \\
\vr_2(t) &= (\cos 2t,\sin 2t),
& t\in[0,\p) \\
\vr_3(t) &= \left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right),
& t\in\RR
\end{align*}
(The last example is @Doug M's from the comments.)
It is a straightforward exercise to show that
\begin{align*}
\vr_1(s) &= (\sin s,\cos s),
& s\in[0,2\p) \\
\vr_2(s) &= (\cos s,\sin s),
& s\in[0,2\p) \\
\vr_3(s) &= \left(\cos s,\sin s\right),
& s\in(-\p,\p)
\end{align*}
Here we have let $t_0=0$ for each parametrized curve for simplicity.
Note that
$\vr_1(-s+\p/2) = \vr_2(s)$.
Under this transformation, $I_1=[0,2\p)\To[-3\p/2,\pi/2)$.
Using periodicity, $I_1\To[0,2\p)$.
Thus, the first and second parametrized curves correspond to the same graph.
Using periodicity we find
$I_3=(-\p,\p)\To[0,2\p)\setminus\{\p\}$.
Thus, the third parametrized curve corresponds to a different graph.
As an exercise, using this procedure one should be able to determine that the following curves correspond to the same graph.
\begin{align*}
&(\sin t,\cos t),
& t\in[0,\p] \\
&(\cos 2t,\sin 2t),
& t\in[-\p/4,\p/4] \\
&\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right),
& t\in[-1,1]
\end{align*}
Best Answer
If $\gamma_2\bigl(\varphi(t)\bigr)=\gamma_1(t)$, then $\gamma_1\bigl(\varphi^{-1}(t)\bigr)=\gamma_2(t)$. However, without the assumption that you always have $\varphi'(t)\ne0$, $\varphi^{-1}$ would not be differentiable, and therefore two parametrizations of the same curve being equivalent would not be an equivalence relation. But it is natural to expect that it is.