Does the vanishing of the Lie derivative of the metric at a single point imply that the associated flow is isometric at some point? Let $(M,g)$ be a smooth Riemannian manifold, and let $X \in \Gamma(TM)$ be a smooth compactly supported vector field on $M$.
Suppose that $(L_X g)(p)=0$ for some specific point $p \in M$. Let $\phi_t$ be the flow of $X$. Is it true that for every $t$, $(d\phi_t)_{q(t)}$ is an isometry for some suitably chosen point $q(t)$? Is it true for $q(t)=p$?

The point is that if we know that $L_Xg=0$ everywhere, i.e. $X$ is Killing, then $\phi_t$ is a global isometry. However, from inspecting the proof, it does not seem "localizable" (i.e. I think that the vanishing of $L_Xg$ at a point should not imply that the flow is an isometry, not even at a single point. But I don't know how to construct an example.)
 A: Note: This answer only addresses the "local" version where we ask if $\mathcal{L}_Xg(p)=0$ implies $\phi_t^*g(q)=g(q)$ for some $p$ in a neighborhood of $p$ and sufficiently small $t$. For the global version, vector fields (esp. those with compact support) may be Killing at some point for entirely unrelated reasons, e.g. the circle has no nowhere-Killing vectors for essentially topological reasons.
Roughly speaking, we can't work pointwise since the flow of $X$ can take us to other points where $\mathcal{L}_X g$ is not zero, and these will contribute to higher order terms in the Taylor expansion of $\phi_t^* g$. We can construct a counterexample with this in mind.
Consider $\mathbb{E}(1)$, i.e. the real line with the standard metric. Let $f$ be a smooth function satisfying the following:
$$\begin{align}
f(0)>0\ \ & \\
f'(0)=0\ \ & \\
0< f'(x)\le 1\ \ &:\ \ x\in\mathbb{R}\setminus\{0\}
\end{align}$$
Now identify $f$ with a vector field on $\mathbb{E}(1)$, (i.e. $f\sim f\partial_1$) and the look at the flow of $f$. The choices above were made so that the infinitesimal flow of $f$ looks like a a infinitesimal translation plus scaling everywhere, with the scale factor positive except at $0$, at which it is zero. The flow will "look like" a local isometry at zero, but will immediately be pushed away from $0$ and acquire a positive scaling factor.
Killing's equation is exceedingly simple in one dimension (since we can identify $(0,2)$ tensors with smooth functions), given by
$$
\left(\mathcal{L}_fg\right)(x)=2f'(x)
$$
Which is zero at $x=0$. However, from the definition of the Lie derivative, the the pullback metric with respect to the (backwards) flow satisfies the differential equation
$$
\frac{d}{dt}\left(\phi_{-t}^*g\right)(x)=2f'(\phi_t(x))
$$
We don't need to solve this equation to see that it is nonzero: For any $T>0$, the left side of the equation is strictly positive except for at most one point in $[0,T]$, and so its integral $\phi_{-T}^*g-g$ is strictly positive, and the flow is not an isometry.
