Is there a closed smooth manifold $M$ such that for each real $x\neq 0$ there is a nowhere vanishing vector field $v$ on $M$ and a diffeomorphism $\phi:M\to M$ such that $\phi_*v=xv$?
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3$\begingroup$ Do you want $v$ to be non-vanishing everywhere? Otherwise you can take $M$ to be the Riemann sphere $\mathbb{C} \cup \{\infty\}$, $v$ to be a multiple of the radial vector field vanishing at $0$ and $\infty$, and the diffeomorphism $\phi$ to just be multiplication by a real number $\lambda > 0$. To get negative constants, I believe you can just compose this with $z \mapsto z^{-1}$. $\endgroup$– Rohil PrasadCommented Sep 10, 2020 at 15:47
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$\begingroup$ @RohilPrasad yes, thank you $\endgroup$– user164740Commented Sep 10, 2020 at 15:57
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Such a manifold exists. First let's construct a non-compact example.
Take $PSL(2,\mathbb R)$ and take two $1$-parameter subgroups, given by $$\begin{pmatrix} e^{t} & 0 \\ 0 & e^{-t} \end{pmatrix}, \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}$$ Consider actions on $PSL(2,\mathbb R)$ of these two groups by multiplication on the left. Then $v$ is the vector field tangent to the second flow, while the first flow will give you a $1$-parameter family of diffeos that will dilate $v$ by any positive constant.
Now, to get the compact example, quotient $PSL(2,\mathbb R)$ from the right by a cocompact action of the fundamental group $\Gamma$ of a compact hyperbolic surface.
It remains to understand how to reverse the sign of $v$. For this recall, that we can identify $PSL(2,\mathbb R)/\Gamma$ with the unit tangent bundle of a hyperbolic surface (whose $\pi_1$ is equal to $\Gamma$). Now, the flow given by $v$ is the horocyclic flow. In order to reverse it, we can take a hyperbolic surface that admits and orientation reversing isometric involution. Such an involution clearly lifts to the unit tangent bundle and it sends the horocyclic flow to its inverse.
answered Sep 10, 2020 at 22:25