Let $h\colon [0,1] \to [0,1]$ a homeomorphism, and $I \colon [0,1]\to[0,1]$ the identity. I want show a homotopy $H:h \sim I$. I want show it in order to show that parametrization of a path is homotopic to that path, but I couldn't explicit it.
We say that $f\colon X \to Y$ and $g\colon X \to Y$ are homotopic if there's a continuous map $H\colon X\times[0,1] \to Y$ such that $H(x,0)=f(x)$ and $H(x,1)=g(x)$. We call $H$ of homotopy from $f$ to $g$.
The standard homotopy when your functional values can be added to each other and multiplied by real numbers is the straight-line homotopy: $$ \begin{align} H:[0,1]\times[0,1]&\to [0,1]\\ (x,t)&\mapsto (1-t)\cdot h(x)+t\cdot I(x) \end{align} $$ It doesn't always work, but it works often enough that you should know it and it should almost always be the first thing you try, in my opinion.
Note that if $h(0)=1$ and $h(1)=0$ (i.e. if $h$ is orientation reversing) then you are better off using the reversed $I$ as well, given by $x\mapsto 1-t$. The homotopy $H$ will behave nicer in that case.