$f$ and $g$ are two arc length parameterisations of a regular simple curve $C$, then prove either $f(s) = g(e+s)$ or $f(s) = g(e-s)\; \forall s$ for some constant $e$.
Intuitively I can understand as the only possible difference between $f$ and $g$ could be the endpoints of the (interval) domain (and both intervals, i.e., domain have the same length). This is true since $\| f' \| = 1 =\| g' \|$.
What I mean is:
$$f: I=[a,b] \rightarrow C$$
$$g: J=[c,d] \rightarrow C$$
$$l(I) = l(J)$$
$$\text{Arc len} = s(t) = \int_a^t \| f' \|ds = \int_c^t \| g'\|ds$$
For the proof, I am thinking in terms of defining $\phi: I \rightarrow J$ followed by using chain rule and fundamental theorem of calculus. But I can't seem to write a formal proof.