This is a sort of draft: I think that since the usual obstruction for constructing smooth conjugation is missing here, this method will work. However, I'm a little bit stuck with final part right now, so either I will add it later or I will delete the answer.
I'm not sure that formulas will be nice, but the construction of conjugating homeomorphism can be done this way (and as far as I remember this is the most simple way to construct this homeomorphism).
Let's start. Take two segments $\lbrack - \varepsilon_U, +\varepsilon_U \rbrack$ and $\lbrack - \varepsilon_V, +\varepsilon_V \rbrack$. Let's build up this mapping $H$ step by step. First, assign $H(-\varepsilon_U) = -\varepsilon_V$ and $H(\varepsilon_U) = \varepsilon_V$. Now observe, that $\lbrack -\varepsilon_U, 0)$ and $(0, \varepsilon_U\rbrack$ are really the $\bigcup\limits_{t\geqslant 0} \varphi^{t}(-\varepsilon_U)$ and $\bigcup\limits_{t\geqslant 0} \varphi^{t}(\varepsilon_U)$ — this is just because $0$ is an attracting hyperbolic equilibrium and we are dealing with 1D system. The same holds for $\lbrack -\varepsilon_V, 0)$ and $(0, \varepsilon_V\rbrack$. Knowing all that we define $H$ the following way. Take a point $x \in \lbrack -\varepsilon_U, 0)$; there exists some unique $\tau_x$ such that $\varphi^{\tau_x}(-\varepsilon_U) = x$. Define $H(x) = H(\varphi^{\tau_x}(-\varepsilon_U)) = \psi^{\tau_x} \Bigl ( H(-\varepsilon_U ) \Bigr ) = \psi^{\tau_x}(-\varepsilon_V)$. Let's check whether this mapping is conjugacy. It is a bijection between $\lbrack -\varepsilon_U, 0)$ and $\lbrack -\varepsilon_V, 0)$. Take any point $y = \varphi^{\tau_y} (-\varepsilon_U)$ and any $t \geqslant 0$. We want to make sure that $\psi^t \circ H$ and $H \circ \varphi^t$ are the same mappings:
$$ \psi^t \Bigl (H(y) \Bigr ) = \psi^t \Biggl (H \Bigl(\varphi^{\tau_y}(-\varepsilon_U) \Bigr) \Biggr ) = \psi^t \Bigl(\psi^{\tau_y}(-\varepsilon_V) \Bigr) = \psi^{t + \tau_y}(-\varepsilon_V) $$
$$ H \Bigl ( \varphi^t(y) \Bigr ) = H \Biggl ( \varphi^t \Bigl (\varphi^{\tau_y}(-\varepsilon_U) \Bigr) \Biggr ) = H \Bigl ( \varphi^{t+\tau_y}(-\varepsilon_U) \Bigr ) = \psi^{t+\tau_y}(-\varepsilon_V)$$
Thus $H$ is really a conjugacy. We can repeat the procedure for constructing $H$ from $( 0, \varepsilon_U \rbrack$ to $(0, \varepsilon_V \rbrack$ and then we will get $H \colon (-\varepsilon_U, +\varepsilon_U) \setminus \{0\} \mapsto
(-\varepsilon_V, +\varepsilon_V) \setminus \{0\}$. It's very natural to set $H(0) = 0$; moreover, we have to do so because we want $H$ to be continuous in the first place. By continuity, if we knew $H(0)$, then after letting $T \rightarrow \infty$ and remembering that $H(\varphi^{T}(-\varepsilon_U)) = \psi^{T}(-\varepsilon_V)$ we get that $H(\varphi^{T}(-\varepsilon_U)) \rightarrow 0$ and
$ \psi^{T}(-\varepsilon_V) \rightarrow 0 $. So, at least $H \colon (-\varepsilon_U, +\varepsilon_U) \mapsto
(-\varepsilon_V, +\varepsilon_V) $ is a nice homeomorphism.
Note that because $\varphi^t$, $\psi^t$ and $\tau_x$ are all diffeomorphisms: when we don't consider any neighbourhood of point $0$ we have that $H$ is just a composition of diffeomorphisms, which of course is a diffeomorphism itself. So we definitely know that $H'(x)$ exists everywhere on $(-\varepsilon_U, +\varepsilon_U)$ maybe except at $0$. I think that for this case it would be enough to prove that $$\lim\limits_{x \rightarrow 0-0} H'(x) = \lim\limits_{x \rightarrow 0+0} H'(x).$$ Using that $$ H \circ \phi^t = \psi^t \circ H$$ we can differentiate both sides, and this is where I am stuck right now.
