The topological conjugacy $h$ guaranteed by the Hartman-Grobman theorem is generally not a lipeomorphism Yesterday I asked a question here Topological conjugacy between linear maps and https://math.stackexchange.com/users/349785/paul-frost answered my question very well ! Thank him. Now I want to ask a very interesting question, at first I prefer to write detail again :
Let $A_{\alpha}: \mathbb{R} \to \mathbb{R}$ denote the linear map
\begin{align}
A_{\alpha}(x)=\alpha x
\end{align}
We proved that if $ 0 &lt; \alpha &lt; 1$ and $ 0 &lt; \beta &lt; 1 $ then $A_{\alpha}$ and $A_{\beta}$ are topologically conjugate.

Notice that two homeomorphisms $f:X \to X$ and $g:X \to X $ are topologically conjugate to each other if there is a homeomorphism $h:X \to X $ such that $hf=gh$

Now the question is this :
What if $\alpha \neq \beta$? In the book of Lan Wen I studied and found out that if $\alpha \neq \beta$ then there is no lipeomorphism $h:\mathbb{R} \to \mathbb{R}$ such that $hA_{\alpha}=A_{\beta}h$. Could anyone help me show this?

Note that We call a homeomorphism $f: E \to E^{'}$ a lipeomorphism if both $f$ and $f^{-1}$ is lipschitz.
Assume that $hA_{\alpha}=A_{\beta}h$ with
$$
c|x-y|\le|h(x)-h(y)|\le d|x-y|.
$$
Then $hA_{\alpha}^n=A_{\beta}^nh$ for all integers $n$ and so
$$
c\alpha^n|x-y|\le|h(A_{\alpha}^nx)-h(A_{\alpha}^ny)|=|A_{\beta}^nh(x)-A_{\beta}^nh(y)|\le d\beta^n |x-y|.
$$
In particular, we have
$$
c\alpha^n|x-y|\le d\beta^n |x-y|.
$$
Now you just need to let $n\to\pm\infty$ to get a contradition.
Assume that $hA_{\alpha}=A_{\beta}h$ with
$$
c|x-y|\le|h(x)-h(y)|\le d|x-y|.
$$
Then $hA_{\alpha}^n=A_{\beta}^nh$ for all integers $n$ and so
$$
c\alpha^n|x-y|\le|h(A_{\alpha}^nx)-h(A_{\alpha}^ny)|=|A_{\beta}^nh(x)-A_{\beta}^nh(y)|\le d\beta^n |x-y|.
$$
In particular, we have
$$
c\alpha^n|x-y|\le d\beta^n |x-y|.
$$
Now you just need to let $n\to\pm\infty$ to get a contradition.