Consider a family of dynamical systems generated by equations: $y'=ax+b, \ \ a,b \in \mathbb{R}$.
Is it true that in this family:
1) There are 4 types of phase portraits up to topological conjugation?
2) $x'=at+b_1 $ and $x'=at+b_2$ are always topologically conjugate.
Here $x=\frac{at^2}{2} + b_1t+C$ and $x=\frac{at^2}{2} + b_2t+C$, $C \in \mathbb{R}$
Two dynamical systems $\varphi: \mathbb{R} \times X \rightarrow X$, $\psi: \mathbb{R} \times Y \rightarrow Y$ ( $X, Y$ - metric spaces) are topologically conjugate if there exists a homeomorphism $h: X \rightarrow Y$ such that $\forall t \in \mathbb{R}, y \in X: \ \psi(t, h(y)) = h(\varphi(t,y))$.
By the way, 3) Does this mean that $h(s)=2s$ is the homeomorphism for systems generated by $y'=y$, $y'=2y$? In the first equation $y=C \cdot e^{x}$, in the second one $y=C \cdot e^{2x}$?
For every $b$, let $$\varphi_b:(t,x)\mapsto\varphi_b(t,x)=\tfrac12at^2+bt+x$$ denote the flow associated to the differential equation $y'(t)=at+b$. That is, for every $(t,x)$, the solution of $y'(t)=at+b$ with initial condition $y(0)=x$ is given by $$y(t)=\varphi_b(t,x). $$ Let us assume that $\varphi_0$ and $\varphi_b$ are conjugate. Thus, there exists some function $h$ such that $$\varphi_b(t,h(x))=h(\varphi_0(t,x)),$$ that is, $$\tfrac12at^2+bt+h(x)=h(\tfrac12at^2+x)\qquad(\ast)$$ for every $(t,x)$. In particular, for $t=1$ and $t=2$, $(\ast)$ yields $$h(x+\tfrac12a)=h(x)+\tfrac12a+b\qquad(\ast\ast) $$ and $$h(x+2a)=h(x)+2a+2b.$$ On the other hand, $(\ast\ast)$ iterated four times yields $$h(x+4\cdot\tfrac12a)=h(x)+4\cdot(\tfrac12a+b).$$ Comparing these two expressions for $h(x+2a)$, one sees that $b=0$. Conversely, this proves that, for $b\ne0$, $\varphi_0$ and $\varphi_b$ are not conjugate.
This approach can be adapted to show that $\varphi_b$ and $\varphi_c$ are not conjugate unless $c=b$.
Turning to the differential equations $y'=ay$, one sees that, now, $$\varphi_a:(t,x)\mapsto\varphi_a(t,x)=x\mathrm e^{at}$$ for every $a$ hence, for every $x\gt0$, the function $$t\mapsto\varphi_a(t,x)^{b/a}=x^{b/a}\mathrm e^{bt}$$ solves $y'=by$ with the initial condition $y(0)=x^{b/a}$. This shows that, for every $(a,b)$, $\varphi_a$ and $\varphi_b$ are conjugate on the domain $(0,+\infty)$, a conjugation function being $$h:z\mapsto h(z)=z^{b/a}.$$ Furthermore, if $ab\gt0$, $\varphi_a$ and $\varphi_b$ are conjugate on the whole real line, a conjugation function being $$h:z\mapsto h(z)=\left\{\begin{array}{ccc}z^{b/a}&\text{if}&z\gt0,\\ 0&\text{if}&z=0,\\ -(-z)^{b/a}&\text{if}&z\lt0.\end{array}\right.$$