This is probably a very elementary question. Nevertheless I decided to post it on MO. Consider a smooth manifold $M$ and a smooth complete vector field $v:M\rightarrow TM$. Consider an autonomous differential equation defined by $v$: $\frac{d}{dt}x(t)=v(x(t)$. Let $\Phi_t(\cdot):M\rightarrow M $ be the evolution map corresponding to time $t$. That is $\Phi_t(x)$ is the solution of differential equation with the initial condition $x\in M$ after time $t$.

## Question

Assume that we know that $\Phi_t(\cdot)$ has at least one fixed point in $M$ for each $t\in\mathbb{R}_+$ (or even $t\in\mathbb{R}$). Under what conditions imposed on the manifold $M$ and a vector field $v$ one can be sure that there is a stationary point of the vector field $v$ in $M$?

## Counterexample

Let $M= \{(x,y) \in\mathbb{R}^2 |1< \sqrt{x^2+y^2} <2 \}$ and let the vector field $v$ has in cylindrical coordinates $(r,\phi)$ the following form: $v=\frac{r-1}{2-r}\frac{\partial}{\partial \phi}$. It is clear that for each $T>0$ we have a circle of periodic orbits with period T. Therefore $\Phi_T(\cdot)$ has fixed points that do not correspond to stationary points of the vector field $v$.

## Modified counterexample

I above example it is sufficient to consider $v=(r-1)\frac{\partial}{\partial \phi}$