Is a general smooth rescaling of a complete vector field itself complete? $\newcommand{\Ga}{\Gamma}$
$\newcommand{\R}{\mathbb{R}}$
$\newcommand{\til}{\tilde}$
$\newcommand{\M}{M}$
$\newcommand{\ep}{\epsilon}$
$\newcommand{\brk}[1]{\left(#1\right)} $
$\newcommand{\R}{\mathbb{R}}$
$\renewcommand{\pd}[2]{\frac{\partial#1}{\partial#2}}$
Let $M$ be a smooth manifold, $X \in \Ga(TM) $.
Assume $X$ is complete, i.e, the flow of $X$ is defined on whole $\mathbb{R} \times M$.
I wonder what happens to the flow when $X$ is multiplied by some real positive function  $f \in C^{\infty}(M)$. My guess is that the flow will still be defined for any time, and that it will be a reparametrization of the original flow. (i.e only the speed may change). In particular $fX$ will be complete.

Question:
  Is this guess correct? Does it hold for non-compact manifolds? (Note I assume anyway $X$ is complete)

Update: As shown by Travis, when $M$ is non-compact, the scaled field need not be complete. Of course, when $M$ is compact, than any vector field is complete. 
For the compact case, I am still interested to know if there is
a global smoothly changing reparametrization $h:\mathbb{R} \times M \to \mathbb{R} $ such that $\psi(t,p)=\phi(h(t,p),p) \forall t \in \mathbb{R} , p \in M$? 
My analysis (below) shows that if there is such a reparametrization than it's unique (since it's enough to prove uniqueness locally), but I do not know how to show there exists such a global $h$.
(See my analysis for details about where my "procedure" gets stuck).

My analysis so far:
Let $\phi_p(t)=\phi(t,p)$ denote the $t$-time flow of $X$ from $p \in M$, i.e
$(1)\,\, \phi: \R \times M \to M \, , \, \dot \phi_p(t)=X(\phi_p(t))$
Take $Y = fX$. Denote the flow of $Y$ by $\psi_p(t)$. Assume there exists a real function $h_p:\R \to \R$ such that $\psi_p(t)=\phi_p(h_p(t))$.
Then $\dot \psi_p(t)=Y(\psi_p(t)) \Rightarrow \dot \phi_p(h_p(t))\cdot h_p'(t)=f(\psi_p(t)) \cdot X(\psi_p(t))$
So by $(1)$ we get: $$X(\psi_p(t)) \cdot h_p'(t)=X\Big(\phi_p\big(h_p(t)\big)\Big)\cdot h_p'(t)=f(\psi_p(t)) \cdot X(\psi_p(t))$$
So if $ X(\psi_p(t)) \neq 0$, this forces $h_p'(t)=f(\psi_p(t))=f\Big(\phi_p\big(h_p(t)\big)\Big)$
This motivates we try to analyze the following equation, $\forall p \in M$:
$$(2) \,\, h_p(0)=0,h_p'(t)=f\Big(\phi_p\big(h_p(t)\big)\Big)$$
We now change notations:
Define $h:\R \times M \to \R$ via: $h(t,p)=h_p(t)$.
Denote $\til M = \R \times M$, and consider the hypersurface $S = \{0\} \times M \subseteq \til M$.
 then $(2)$ becomes:
$$(3) \, \, h|_S=0, \pd{}{t} h = (f \circ \phi) \big(h(t,p),p\big)$$
The vector field $\pd{}{t} \in \Ga\brk{T \til \M}$ is nowhere tangent to $S$ (since $T_{\brk{0,p}}S=0 \oplus T_p\M$ and $\pd{}{t}(t,p)=(1,0)$).
Denote $C^\infty\brk{\til M \times \R} \ni \til f: \til M \times \R \to \R$ via the formula:
$$\til f((t,p),s) = (f \circ \phi)(s,p) $$, then $(3)$ becomes:
$$ (4) \, \,  h|_S=0, \pd{}{t} h = \til f \big((t,p),h(t,p)\big)$$
The above equation is an instance of a Quasilinear Cauchy problem (on the manifold $\til M$), so we know $\forall \til p=(0,p) \in S$ there exists a unique solution in some neighbourhood $U$ of $\til p$.
(See for instance Theorem 9.53, page 242 in John M.Lee's book "Introduction to smooth manifolds")
In the case $M$ is compact, we can proceed in the following way:
$\forall p \in \M , (0,p) \in S \Rightarrow  (0,p) \in U \Rightarrow$ there exists an open set $\til U_p \subseteq U$ which contains $(0,p)$. Hence, there exists $\ep_p \in \R \, , \, U_p \subseteq \M$ ($U_p$ open in $\M$)  such that $(-\ep_p,\ep_p) \times U_p \subseteq \til U_p$. $\{U_p|p \in \M\}$ form an open cover of $\M$, hence (by compactness of $M$) there is a finite subcover $U_{p_1},\dots ,  U_{p_n}$.
Define $\ep = min\{\ep_{p_i}|i=1,\dots,n \}$. It follows immediately that $(-\ep,\ep) \times \M \subseteq U$.
So, we have stablished exsitence of a unique solution on $(-\ep,\ep) \times \M$.
The problem is how to continue from here.
A naive approach is to define
 $X = \{t \in I| \text{there exists a unique solution for $(4)$ in  } (-t,t) \times \M \}$.
Look at $s= \text{sup} X$. We claim $\text{sup} X \in X$. Since $X$ is closed downward, i.e :  $x \in X \Rightarrow \brk{0 \le x' < x \Rightarrow x' \in X}$ it follows that $[0,s) \subseteq X$.
It's easy to see there must be a unique solution for $(4)$ on $(-s,s) \times \M$. (If there were two different solutions, they would differ already at some $s'<s$ which contradicts $[0,s) \subseteq X$).
Hence, by continuity, there is at most one solution on $[-s,s] \times \M$.
So, if we knew it's possibe to extend the unique solution on $(-s,s) \times \M$ to $[-s,s] \times \M$, then we could advance the (existence and) uniqueness further, via the same argument, thus obtaining a contradiction.
(In that case our initial hypersurfaces would have been $\{s\} \times M$,$\{-s\} \times M$).
I am not sure how to show the solution can be extended that way.
In principle, we can also start the equation from $t=s$ by requiring $h(s,p)$ to satisfy $\psi(s,p)=\phi(h(s,p),p)$. Since non-constant integral curves are injective or periodic, we this might  determine $h_{\{s\} \times M}$ uniquely and perhaps we can continue from there.
(Although then the issue of smoothness arise).
 A: For general (and in particular, noncompact) manifolds this is false:
Hint Consider the standard coordinate vector field $$X := \partial_x$$ on $\Bbb R$, which is complete (its flow is $(t, a) \to a + t$, which in particular is defined on all of $\Bbb R \times \Bbb R$), and take $$f(x) := 1 + x^2 . $$

 Then, substituting in the definition gives that the flow curve $\psi_0(t)$ of $f X$ starting at the origin is the solution to the i.v.p. $$\psi_0'(t) = \left[1 + \psi_0(t)^2\right] \partial_x, \qquad \psi_0(0) = 0.$$ Writing the sole component in the global frame $(\partial_x)$, applying the Method of Separation by Variables, and carrying out a straightforward integration gives that the flow curve is $$\psi_0(t) = \tan t, \qquad -\tfrac{\pi}{2} < t < \tfrac{\pi}{2},$$ which in particular cannot be extended even continuously outside this interval. So, $f X$ is not complete.

Of course, any smoothly rescaled smooth vector field $f X$ is still smooth, so its flow $\psi_a(t)$ starting at any point $a$ exists for some finite interval $(-\epsilon, \epsilon)$, and it shouldn't be too much work to write down a reasonably explicit lower bound for $\epsilon$ for any $\delta > 0$ in terms of $\delta$ and an upper bound for $|f|$ on $(a - \delta, a + \delta)$ (and, to be clear, on $a$).
On the other hand, the conjecture in the question is true if we add one of two simple hypotheses: It follows from the dramatic-sounding Escape Lemma that if $X$ is complete and


*

*$|f|$ is bounded and/or

*$M$ is compact,


then $f X$ is complete. (In this latter case, this is for the simple reason that any smooth vector field on a compact manifold is complete.)
