Here is Theorem 6.20 in the book Mathematical Analysis - A Modern Approach to Advanced Calculus by Tom M. Apostol, 2nd edition:
Let $\mathbf{f} \colon [a, b] \longrightarrow \mathbb{R}^n$ and $\mathbf{g} \colon [c, d] \longrightarrow \mathbb{R}^n$ be two paths in $\mathbb{R}^n$ (i.e. continuous functions), each of which is one-to-one on its domain. Then $\mathbf{f}$ and $\mathbf{g}$ are equivalent (i.e. $\mathbf{g} = \mathbf{f} \circ u$ for some function $u \colon [c, d] \longrightarrow [a, b]$ such that $u$ is continuous, strictly monotonic, and onto) if and only if they have the same graph (i.e. range).
The proof Apostol gives is as follows:
Equivalent paths necessarily have the same graph. To prove the converse, assume that $\mathbf{f}$ and $\mathbf{g}$ have the same graph. Since $\mathbf{f}$ is one-to-one and continuous on the compact set $[a, b]$, Theorem 4.29 tells us that $\mathbf{f}^{-1}$ exists and is continuous on its graph. Define $u(t) = \mathbf{f}^{-1} \big[ \mathbf{g}(t) \big]$ if $t \in [c, d]$. The reader can verify that $u$ is strictly monotonic [How to do this?], and hence $\mathbf{f}$ and $\mathbf{g}$ are equivalent paths.
How to show explicitly that $u$ is indeed strictly monotonic?
Of course, the function $u \colon [c, d] \longrightarrow [a, b]$ is continuous, one-to-one, and onto.
If $u$ is not monotone, there exists $x_1 < x_2 < x_3$ such that $f(x_2) = \min \{f(x_1), f(x_2), f(x_3) \}$. Assume $f(x_2) \leq f(x_1), f(x_3)$. Take $c \in [f(x_2), \min \{f(x_1), f(x_3) \}]$. By immediate value theorem if continuous $u$ on $[x_1, x_2]$ and $[x_2, x_3]$, there exists $y_1, y_2$ such that $f(y_1) = f(y_2)$