$\textbf{Problem}$ :
Show directly that if $f, g \in A u t(\mathbb{I}),$ and $$ f^{-1}((f(x)+f(y)-1) \vee 0)=g^{-1}((g(x)+g(y)-1) \vee 0) $$ for all $x, y \in[0,1],$ then $f=g$
$\textbf{Note}$ : $A u t(\mathbb{I})$ is the set of all functions $f$ from [0,1] to [0,1] that are one-to-one and onto, and such that $f(x) \leq f(y)$ if and only if $x \leq y .$
(Getting another question off the unanswered list.)
Suppose that $f,g\in\mathrm{Aut}(\Bbb I)$, and
$$f^{-1}\Big(\big(f(x)+f(y)-1\big)\lor 0\Big)=g^{-1}\Big(\big(g(x)+g(y)-1\big)\lor 0\Big)$$
for all $x,y\in[0,1]$; since $f$ and $g$ are bijections, so are $f^{-1}$ and $g^{-1}$, and therefore
$$\big(f(x)+f(y)-1\big)\lor 0=\big(g(x)+g(y)-1\big)\lor 0$$
for all $x,y\in[0,1]$. Note also that since $f$ and $g$ are order-preserving and surjective, $f(1)=g(1)=1$. Thus,
$$\begin{align*} f(x)&=f(x)\lor 0\\ &=\big(f(x)+f(1)-1\big)\lor 0\\ &=\big(g(x)+g(1)-1\big)\lor 0\\ &=g(x)\lor 0\\ &=g(x) \end{align*}$$
for each $x\in[0,1]$, i.e., $f=g$.