[proofplan]
We show that the flow $g^\varepsilon x := \tilde{x}(\varepsilon)$ of the ODE $d\tilde{x}/d\varepsilon = V(\tilde{x})$ is a one-parameter group with generating vector field $V$. This requires verifying three properties: (1) $g^0 = \operatorname{id}$, which follows from the initial condition; (2) the group law $g^{\varepsilon_1 + \varepsilon_2} = g^{\varepsilon_1} \circ g^{\varepsilon_2}$, which follows from the autonomous character of the ODE and uniqueness of solutions; and (3) the infinitesimal generator is $V$, which follows by differentiating the flow at $\varepsilon = 0$.
[/proofplan]
[step:Verify that $g^0 = \operatorname{id}$ from the initial condition]
By definition, $g^\varepsilon x = \tilde{x}(\varepsilon)$ where $\tilde{x}(\varepsilon)$ solves
\begin{align*}
\frac{d\tilde{x}}{d\varepsilon} = V(\tilde{x}), \qquad \tilde{x}(0) = x.
\end{align*}
Evaluating at $\varepsilon = 0$:
\begin{align*}
g^0 x = \tilde{x}(0) = x.
\end{align*}
Since this holds for all $x$, we have $g^0 = \operatorname{id}$.
[/step]
[step:Establish the group law $g^{\varepsilon_1 + \varepsilon_2} = g^{\varepsilon_1} \circ g^{\varepsilon_2}$ by uniqueness]
Fix $x$ and $\varepsilon_2$, and set $y := g^{\varepsilon_2} x = \tilde{x}(\varepsilon_2)$. Consider two curves parametrised by $\varepsilon_1$:
**Curve 1.** $\alpha(\varepsilon_1) := g^{\varepsilon_1 + \varepsilon_2} x = \tilde{x}(\varepsilon_1 + \varepsilon_2)$. Since $\tilde{x}(\varepsilon)$ solves $d\tilde{x}/d\varepsilon = V(\tilde{x})$, the time-shifted curve satisfies
\begin{align*}
\frac{d\alpha}{d\varepsilon_1} = \frac{d}{d\varepsilon_1} \tilde{x}(\varepsilon_1 + \varepsilon_2) = V(\tilde{x}(\varepsilon_1 + \varepsilon_2)) = V(\alpha(\varepsilon_1)),
\end{align*}
with initial condition $\alpha(0) = \tilde{x}(\varepsilon_2) = y$.
**Curve 2.** $\beta(\varepsilon_1) := g^{\varepsilon_1} y$. By definition of the flow, $\beta$ solves
\begin{align*}
\frac{d\beta}{d\varepsilon_1} = V(\beta(\varepsilon_1)),
\end{align*}
with initial condition $\beta(0) = g^0 y = y$.
Both $\alpha$ and $\beta$ solve the same autonomous ODE $dz/d\varepsilon_1 = V(z)$ with the same initial condition $y$ at $\varepsilon_1 = 0$. Since $V$ is smooth (and hence locally Lipschitz), the Picard-Lindelof theorem guarantees uniqueness, so $\alpha(\varepsilon_1) = \beta(\varepsilon_1)$ for all $\varepsilon_1$. That is,
\begin{align*}
g^{\varepsilon_1 + \varepsilon_2} x = g^{\varepsilon_1}(g^{\varepsilon_2} x)
\end{align*}
for all $\varepsilon_1, \varepsilon_2$ and all $x$, establishing the group law $g^{\varepsilon_1 + \varepsilon_2} = g^{\varepsilon_1} \circ g^{\varepsilon_2}$.
[/step]
[step:Verify that $V$ is the infinitesimal generator of $g^\varepsilon$]
The infinitesimal generator of the one-parameter group $g^\varepsilon$ is, by definition, the vector field obtained by differentiating the action at $\varepsilon = 0$:
\begin{align*}
\left.\frac{d}{d\varepsilon}\right|_{\varepsilon = 0} g^\varepsilon x = \left.\frac{d\tilde{x}}{d\varepsilon}\right|_{\varepsilon = 0} = V(\tilde{x}(0)) = V(x).
\end{align*}
The first equality is the definition of $g^\varepsilon$, the second uses the ODE $d\tilde{x}/d\varepsilon = V(\tilde{x})$, and the third uses the initial condition $\tilde{x}(0) = x$.
Since this holds at every $x$, the infinitesimal generator of $g^\varepsilon$ is the vector field $V$. Combined with the group law and the identity property established above, this shows that $g^\varepsilon$ is a one-parameter group of transformations with generating vector field $V$.
[/step]
