Let $M$ be a space and $I$ the unit interval.
Definition A map $f : I \to M$ is a parametrization by arc-length if $f$ maps $I$ diffeomorphically onto an open subset of $M$, and if the "velocity vector" $df_s(1)$ has unit length, for each $s\in I$.
Now let $f:I\to M$ and $g:J\to M$ be parametrizations by arc-length.
Claim The derivative of $g^{-1}\circ f$ is equal to $\pm 1$ everywhere.
Why is the claim true?
The derivative of $g^{-1}\circ f$ is $dg^{-1}\circ df=∂g^{-1}/∂f^1\cdot∂f^1/∂x$. Since $dg^{-1}$ and $df$ have unit length, $∂g^{-1}/∂f^1=\pm1$ and $∂f^1/∂x=\pm 1$ and the claim follows, I think.
How do we know that $f$ has only one component function? Is it because it is a diffeomorphism between $I$ and $f(I)$?
Note that $g^{-1} \circ f$ does not make sense unless $f$ takes values in the image of $g$.
It is true that the derivative of $g^{-1} \circ f$, which is a real number, has absolute value $1$ everywhere on $I$, so it must be $\pm 1$ everywhere. By continuity of $(g^{-1} \circ f)'$ and connectedness of $I$, it must be either $+1$ everywhere or $-1$ everywhere.
FYI, a careful little computation shows that the derivative of $g^{-1} \circ f$ at $t\in I$ is the unique real number $\alpha(t) := (g^{-1} \circ f)'(t)$ such that $f'(t) = \alpha(t)\,g'(s)$ where $s = g^{-1} \circ f (t)$. Note that in this equality, $f'(t)$ and $g'(s)$ are not real numbers but tangent vectors (at $f(t)$) to $M$.