Let $M$ be a smooth manifold endowed with an action from the Lie group $S^1$ (identified as complex numbers of absolute value $1$) as such: $$\cdot:M \times S^1 \to M, (p,z) \mapsto p \cdot z $$ and such that $p \cdot 1 = p, \forall p \in M$ and $(p \cdot z_1) \cdot z_2 = p \cdot (z_1z_2), \forall z_1, z_2 \in S^1.$ Now define the vector field $V$ such that $$V: p \mapsto V_p, \text{ with } V_p = \frac{d}{dt} \left|_{t=0} \right. \ p \cdot e^{2\pi it}. $$
What is the flow of $V$ and why does $p \cdot z = p , \forall z \in S^1 \iff V_p=0$? I tried looking locally in a chart $(U,\varphi)$ around a point $p \in M$ and specifically finding an integral curve $\gamma_p$ of $V$ (i.e. $\gamma_p(t) = V_{\gamma_p(t)}),$ but I don't know how to get past the action of $S^1$ to specifically solve the equation and find $\gamma_p$.
The flow of $V$ is $\phi_t(x)=x.e^{2\pi it}$ since ${d\over{dt}}_{t=0}\phi_t(x)=V(x)$,
$p=p.z$ implies that ${d\over{dt}}_{t=0}p.e^{2\pi it}p={d\over{dt}}_{t=0}p=0=V(p)$. On the other hand if $V(p)=0$, $\phi_t(p)=p$ for every $t$ which is equivalent to $p.z=p$ for every $z\in S^1$.