Uniqueness is a consequence of a non-vanishing autonomous flux Let $f: \mathbb R\to\mathbb R$ continuous and $f(x)\ne 0$, or all $x\in\mathbb R$. Let also $\varphi,\psi : \mathbb R\to\mathbb R$, continuously differentiable functions satisfying
$$
\varphi'(t)=f\big(\varphi(t)\big),\quad \psi'(t)=f\big(\psi(t)\big),
$$
for all $t\in\mathbb R$. Show that there exists a $\tau\in\mathbb R$, such that
$$
\psi(t)=\varphi(t-\tau),\quad \text{for all $t\in\mathbb R$}.
$$
Note. Here uniqueness is the question, but in order to even consider proving it, one has to first make sure that the ranges of the two solutions intersect! 
 A: This just proves that the ranges must overlap. In fact, 
the range of $\phi$ and $\psi$ is $\mathbb{R}$.
Without loss of generality, we may take $f$ to be positive everywhere. Hence both $\phi,\psi$ are strictly increasing. 
Let $\psi_\infty = \lim_{t \uparrow \infty} \psi(t)$, and suppose $\psi_\infty < \infty$.
 Let $\psi_\infty = \lim_{t \uparrow \infty} \psi(t)$. Since $f$ is continuous, we must have $f(\psi_\infty) = \lim_{t \uparrow \infty} f(\psi(t))$. Since $f(\psi_\infty)>0$, we must have $\lim_{t \uparrow \infty} \psi(t) = \infty$, a contradiction. Hence $\psi_\infty = \infty$.
The other limits follow in exactly the same fashion.
Uniqueness was more straightforward that I thought. It follows from the fact that $f(x) \neq 0$:
Suppose $x_0 = \phi(t_0) $. Let $g(x) = \int_{x_o}^x { dy \over f(y) }$. Note that $g(x_0) = 0$ and $g'(x) = {1 \over f(x) } \ne 0$. The inverse function theorem gives the existence of a function $\gamma$ defined on a neighbourhood $U$ of $0$ such that $\gamma(g(x)) = x$ for $x \in V$, where $V$ is some neighbourhood of $x_0$.
Now consider $f(t) = g(\phi(t))$. Then $f(t_0) = 0$, and $f'(t) = g'(\phi(t)) \phi'(t) = 1$, and so $f(t) = t-t_0$. Hence for $t-t_0 \in U$ we have $\gamma(f(t)) =\gamma(t-t_0) = \phi(t)$.
Suppose $\psi$ is another solution that passes through $(t_0, x_0)$, then the
same considerations show that $\psi(t) = \gamma(t-t_0) = \phi(t)$, for $t-t_0 \in U$. 
Let $T\subset \mathbb{R}$ be the points at which two continuous solutions match. This set is closed by continuity, and open by the above reasoning, hence since $\mathbb{R}$ is connected either $T$ is empty or the whole space.
Since the ranges of $\phi, \psi$ overlap, there are points $t_0,t_1$ such that $\phi(t_0) = \psi(t_1)$. Since the system is time invariant, if $\phi$ is a solution, then so is $t \mapsto \phi(t-s)$ for any fixed $s$. Hence $t \mapsto \phi(t+t_0)$ and $t \mapsto \psi(t+t_1)$ are both solutions that pass through the point $(0,\phi(t_0)) = (0,\psi(t_1))$. Consequently $\phi(t+t_0)=\psi(t+t_1)$ for all $t$, from which we get $\psi(t) = \phi(t+t_0-t_1)$ for all $t$.
