Let $\mathbb{I}=([0,1],\leq)$ and suppose
$Aut(\mathbb{I})=\{f|f:\mathbb{I}\longrightarrow \mathbb{I}, \text{$f$ is 1-1 and onto and $x\leq$ y iff $f(x)\leq f(y)$}\}$.
For any $f\in Aut(\mathbb{I})$ and $r\in \mathbb{R}^+$, define $rf$ by $(rf)(x)=f(x)^r$. I am trying to show that $f^{-1}\mathbb{R}^+f\subsetneqq Aut(\mathbb{I})$ where $f^{-1}\mathbb{R}^+f=\{f^{-1}rf|r\in \mathbb{R}^+\}$ and $(f^{-1}rf)(x)=f^{-1}(f(x)^r)$ for all $x\in [0,1]$. Of course it is obvious that $f^{-1}\mathbb{R}^+f\subseteq Aut(\mathbb{I})$. But why $f^{-1}\mathbb{R}^+f\neq Aut(\mathbb{I})$? Thanks.
We consider $r>0$ as the isomorphism $x\mapsto x^r$, so $rf$ and $f^{-1}rf$ are just compositions. So $\mathbb R^+$ is a (proper) subgroup of $Aut(\mathbb I)$, and $f^{-1}\mathbb R^+f$ are its conjugates, hence also proper subgroups.