The system $x'=Ax$ is an attractor. Let $h$ be defined by
$$h(0)=0 \qquad h(x)=e^{t_x}e^{t_xA}x$$ where $t_x$ is the real number such that $q(e^{t_x A}x)=1$ and $q(x)=\int_0^{\infty}\langle e^{tA}x,e^{tA}x\rangle\,dt$.
The book says that $h(\cdot)$ is continuous and is the topological conjugacy of $x'=Ax$ and $x'=-x$.
Knowing that $t_x$ is $C^{\infty}$ in $\mathbb{R}^n\setminus\{0\}$ I managed to show that $h|_{\mathbb{R}^n\setminus\{0\}}$ is continuous. But I'm not able to show that this function is continuous at 0.
You first should show that there exist $a,b>0$ such that $$ a\|x\|^2\le q(x)\le b\|x\|^2. $$ This follows easily from having an attractor (use the eigenvalues of $A$).
Now observe that $$ \begin{split} q(h(x)) &=\int_0^{\infty}\langle e^{tA}e^{t_x}e^{t_xA}x,e^{tA}e^{t_x}e^{t_xA}x\rangle\,dt\\ &=e^{2t_x}\int_0^{\infty}\langle e^{tA}e^{t_xA}x,e^{tA}e^{t_xA}x\rangle\,dt\\ &=e^{2t_x}q(e^{t_xA}x)=e^{2t_x}. \end{split} $$ Therefore, $$ a\|h(x)\|^2\le e^{2t_x}\le b\|h(x)\|^2. $$ On the other hand, since $$ 1=q(e^{t_xA}x) =\int_0^{\infty}\| e^{(t+t_x)A}x\|^2\,dt= \int_{t_x}^{\infty}\| e^{tA}x\|^2\,dt, $$ we find that if $x\to0$, then $t_x\to-\infty$. It follows from $$a\|h(x)\|^2\le e^{2t_x}$$ that $h(x)\to0$ when $x\to0$. The other inequality can be used for the inverse.