Are the diffeomorphisms $x^3 + \varepsilon$ and $2x$ conjugated? ($\varepsilon > 2/(3\sqrt{3})$) Consider the following two diffeomorfisms
\begin{align*}
f:\mathbb{R}&\to \mathbb{R}\\
x&\mapsto x^3 + \varepsilon,
\end{align*}
where $\varepsilon>2/(3\sqrt{3})$, and
\begin{align*}
g:\mathbb{R}&\to \mathbb{R}\\
x&\mapsto 2x,
\end{align*}
Are the functions $g$ and $f$ conjugated? i.e. I would like to know if there exists a homeomorphism $h:\mathbb{R} \to \mathbb{R}$ such that 
$$h\circ f = g \circ h.$$
Such homeomorphism seems like to exist since these two diffeomorphism have the exact same dynamics, however I do not know how to construct / argue existence of $h$.
Can anyone help me?
 A: Short answer: Yes, they are conjugate. This follows from Theorem $2$ below, which gives a general characterization of when two homeomorphisms of $\mathbb{R}$ are topologically conjugate.
For completeness, I will write down the construction alluded to in https://mathoverflow.net/questions/20246/how-to-construct-a-topological-conjugacy. First we consider the case of no fixed points, and then glue together the this construction applied to the intervals between fixed points when they occur.
Theorem 1. Let $f, g \colon \mathbb{R} \rightarrow \mathbb{R}$ be orientation-preserving (increasing) homeomorphisms with no fixed points. Then $f$ and $g$ are topologically conjugate.
Proof. Let $x_0 \in \mathbb{R}$ and consider its orbit under $f$, namely $\{\ldots, f^{-1}(x_0), x_0, f(x_0), \ldots \} = \{\ldots, x_{-1}, x_{0}, x_{1}, \ldots \}$ where $x_i := f^{i}(x_0)$. Similarly take $x'_{0} \in \mathbb{R}$ and its orbit under $g$, $\{\ldots, x'_{-1}, x'_{0}, x'_{1}, \ldots \}$. Let $h_0$ be an orientation preserving homeomorphism sending $[x_0, x_1]$ to $[x'_0, x'_1]$ (for instance, there's a unique affine, orientation preserving one). Note that because $f$ is increasing, the intervals $[x_i, x_{i+1}]$ partition $\mathbb{R}$ (they only overlap at their endpoints). We can say the same about the intervals $[x'_i, x'_{i+1}]$. So, we will inductively define our conjugacy, $h$, on each of these intervals by using $h_0$. We begin with the intervals with $i \geq 1$. 
Suppose the construction of an increasing, continuous $h_i$ has been done, we define $h_{i+1}$ as follows:
$$ \begin{align} h_{i+1} &\colon [x_{i+1}, x_{i+2}] \rightarrow \mathbb{R}
\\ h_{i+1}(x) &= g \circ h_i \circ f^{-1} (x) \end{align} $$
Now we check that, for $x \geq x_0$, $h$ satisfies the conditions of being a conjugacy. Suppose $x \in [x_j, x_{j+1}]$ for some $j \geq 0$, so that $f(x) \in [x_{j+1}, x_{j+2}]$. Then
$$ \begin{align} h \circ f (x) &= h_{j+1} \circ f(x) \\
&=g \circ h_j \circ f^{-1}(f(x))\\
&=g \circ h_j(x)\\
&=g \circ h(x) \text{ as desired} \end{align} $$
Note that  $h|_{[x_{i+1}, x_{i+2}]} = h_{i+1} = (g \circ h_i)|_{[x_i, x_{i+1}]}$. We assumed that $h_i$ and $g$ was increasing, so $h$ is increasing on each of the intervals $[x_i, x_{i+1}]$, $i \geq 0$. Clearly it's continuous on these intervals. If we show that the $h$ is well defined on the endpoints, we get that it is increasing and continuous for $x \geq x_0$. So, consider $j \geq 1$. We have that $x_{j+1}$ appears in the domain of $h_j$ and $h_{j+1}$. We have that 
$$\begin{align}
h_j(x_{j+1}) &= g(h_{j-1}(f^{-1}(x_{j+1}))) = g(h_{j-1}(x_j)) \\
h_{j+1}(x_{j+1}) &= g(h_j(f^{-1}(x_{j+1}))) = g(h_j(x_j))
\end{align}$$
Thus the construction being well-defined at the end points depends on the base case, so we only need to check that $h_0(x_1) = h_1(x_1)$. We have $h_0(x_1) = x'_1$ by construction of $h_0$. Also by construction, we have $h_1(x_1) = g(h_0(f^{-1}(x_1))) = g(h_0(x_0)) = g(x'_0) = x'_1$. So far we have shown that $h$ is a topological conjugacy between $f$ and $g$ on $[x_0, \infty)$.
To extend $h$ on the intervals $[x_i, x_{i+1}]$ for $i \leq -1$ is a very similar process. Suppose $h_i$ is defined for some $i \leq 0$. This time we define
$$ \begin{align} h_{i-1} &\colon [x_{i-1}, x_{i}] \rightarrow \mathbb{R}
\\ h_{i-1}(x) &= g^{-1} \circ h_i \circ f (x) \end{align} $$
To prove this is a conjugacy on those intervals is very similar to the previous case. For instance, let $x \in [x_{j-1}, x_j]$ with $j \leq 0$. Then $$g \circ h(x) = g \circ h_{j-1} (x) = g(g^{-1} \circ h \circ f (x)) = h \circ f(x).$$ We leave the other details to the reader. This proves that $h$ is a conjugacy between $f$ and $g$.
Corollary (Your Case). Let $f, g \colon \mathbb{R} \rightarrow \mathbb{R}$ be orientation-preserving homeomorphisms, both with only a single fixed point and that fixed point being repelling. Then $f$ and $g$ are topologically conjugate.
Proof. Let $z$ be the fixed point of $f$ and $z'$ the fixed point of $g$. Using Theorem $1$ we can construct partial conjugacies between $f$ and $g$ on the appropriate intervals:
$$\begin{align}
h_{a} &\colon (-\infty, z) \rightarrow (-\infty, z') \\
h_{b}&\colon (z, \infty) \rightarrow (z', \infty) \\
\end{align}$$
If $h$ is to be a conjugacy, we are forced to have $h(z) = z'$, so we define $h_{a}(z) = z'$ and $h_{b}(z) = z'$. This extension makes $h$ continuous (this requires some checking, not hard but a bit annoying).
This leads us to the following natural generalization:
Theorem 2. Let $f,g \colon \mathbb{R} \rightarrow \mathbb{R}$ be two orientation-preserving homeomorphisms. Then $f$ and $g$ are topologically conjugate if and only if they have the same number of fixed points of each type (repelling, attracting, neutral) and the types of fixed points appear either in the same order or in reverse order.
Proof. Suppose that $f,g$ have the same number of each type of fixed point and both sets of them appear in order. Then we can inductively construct the conjugacy in the same manner as in the corollary above. If they appear in reverse order, then we must make the conjugacy decreasing. We can do this by using nearly the same construction, only changing $h_0$ in Theorem $1$ to a decreasing homeomorphism. Now suppose that some $3$ fixed points appear out of order. That is to say $f$ has some fixed points $a < b < c$ with $b$ being of a different type than $a$ and $c$, and that $g$ has some fixed points $a' < c' < b'$ (WLOG, this could also be $b' < a' < c'$). Suppose that $a$ is of the same type as $a'$, and so on. If such a triple of points exists for $f$, then we can take it to be one such that the leftmost element (namely $a$) is the largest possible, or that the rightmost element (namely $c$) is the smallest possible. This is a small technical detail: it forces the image of a conjugacy $h$ on $a$, $b$, and $c$ to be the largest possible triple, or the smallest possible such triple, otherwise we'd contradict monotonicity of $h$. Thus we have $a < b < c$ and $h(a) < h(c) < h(b) = b'$ (or the same with the inequalities reversed, if $h$ were to be decreasing). This contradicts monotonicity of $h$. Thus, no such homeomorphism can exist. 
