TITLE: Relation between the eigenvalues of matrices conjugated by a homeomorphism.
QUESTION [5 upvotes]: Let $A, B$ be $2\times 2 $ matrices satisfying:

The eigenvalues $\lambda,\mu$ of  $A$ satisfy $|\lambda|<1<|\mu|$. 
The eigenvalues $\lambda',\mu'$ of  $B$ satisfy $|\lambda'|<1<|\mu'|$.
There exists a homeomorphism $h:\mathbb{R}^2 \to \mathbb{R}^2$ such that $$h(A x) = B h(x), \  \forall \ x \in \mathbb{R}^2.$$

I'm reading the paper "Generic Singularities of 3D Piecewise
Smooth Dynamical Systems", and the author "kinda says" that that 1) + 2) + 3) implies that 
$$\frac{\log(\lambda)}{\log(\mu)} =\frac{\log(\lambda')}{\log(\mu')}.  $$

My question: Does someone know if 1)+2)+3) $\Rightarrow$ $\frac{\log(\lambda)}{\log(\mu)} =\frac{\log(\lambda')}{\log(\mu')} $? 


Just some commentaries.
I might be confusing something. However, my assumption is based on this phrase 

The references listed on the picture above are:

Moreover, on Proposition 9, the author uses (in my view) the fact of existing a conjugation between two diffeomorphisms $\phi$ and $\phi_0$, to conclude that $P(\phi) = P(\phi_0)$ and then construct a homeomorphism.
Can anyone help me?

EDIT: the author at no time says that $1) + 2) +3) \Rightarrow \frac{\log(\lambda)}{\log(\mu)} =\frac{\log(\lambda')}{\log(\mu')}$, I understood what was written in the wrong way, it was my mistake.

REPLY [2 votes]: This is not true. Consider the linear maps $A(x,y)=(\frac12x,\,2y),\ B(x,y)=(\frac12x,\,8y)$ and the function $h(x,y)=(x,y^3)$ defined on $\mathbb R^2$. Then $h$ is a homeomorphism and
$$
h(A(x,y))=h\left(\frac12x,\,2y\right)=\left(\frac12x,\,8y^3\right)
=B(x,\,y^3)=B(h(x,y)),
$$
but
$$
\frac{\log(\lambda)}{\log(\mu)}=\frac{\log(\frac12)}{\log(2)}
\ne\frac{\log(\frac12)}{\log(8)}=\frac{\log(\lambda')}{\log(\mu')}.
$$