Killing Fields on Euclidean Spaces

We need the following lemma, whose proof will be at the bottom of this answer.

Lemma: If $\phi, \psi$ are Riemannian isometries on open subsets $U \to V$ of $\mathbb R^n$, with $\phi(p) = \psi(p)$ and $d\phi_p = d\psi_p$ for some $p \in U$, then $\phi = \psi$.

Now, suppose $V$ is a Killing vector field vanishing at $0$ in the kernel of the map $T : K \to \mathfrak o(n)$. Since $V$ vanishes at $0$, if $\theta$ is the flow of $V$, $\theta_t(0) = 0$ for every $t$ in the flow domain. We claim $d(\theta_t)_0$ is independent of $t.$ For then, by the lemma, $\theta_t = \mathrm{Id}$ on all of $\mathbb R^n$, and so $V = 0$.

Let $\theta_t^i$ be the $i^\textrm{th}$ component of the flow $\theta_t$. Then $\dfrac{d}{dt}\bigg|_{t=0} \theta^i_t(x) = V^i(x)$ for $x \in \mathbb R^n$, where $V^i$ are the component functions of $V$. To prove the claim, it suffies to show that $\dfrac{\partial \theta_t^i}{\partial x^j}(0)$ is independent of $t$ for every $i,j$. This follows from Clairaut's theorem: for every $t_0$ in the flow domain, we have $$ \frac{d}{dt}\bigg|_{t=t_0} \frac{\partial}{\partial x^j} \theta_t^i(0) = \frac{\partial}{\partial x^j} \frac{d}{dt}\bigg|_{t=t_0} \theta_t^i(0) = \frac{\partial}{\partial x^j} \frac{d}{dt}\bigg|_{t=0} \theta^i_t(\theta_{t_0}(0)) = \frac{\partial}{\partial x^j} V^i(\theta_{t_0}(0)) = \frac{\partial V^i}{\partial x^j}(0) = 0 $$ since $V \in \ker(T)$. Therefore $\dfrac{\partial \theta_t^i}{\partial x^j}(0)$ is independent of $t$ for every $i,j$, hence so is $d(\theta_t)_0$, so $V=0$ by the lemma, so $T$ is injective. QED.

Proof of lemma: Assume $U$ is convex. Isometries take line segments to line segments, so $V$ is also convex. For $q \in U$, let $\gamma(t) = p + t(q-p)$. Then $\phi \circ \gamma$ and $\psi \circ \gamma$ are both line segments from $r := \phi(p) = \psi(p)$ to $\phi(q)$ and to $\psi(q)$ respectively; that is, $\phi \circ \gamma(t) = r + t(\phi(q)-r)$ and $\psi \circ \gamma(t) = r + t(\psi(q) - r)$. Since $d\phi_p = d\psi_p$, we get $$\phi(q) - r = \frac d{dt}\bigg|_{t=0} (\phi \circ \gamma)(t) = d\phi_p(\dot\gamma(0)) = d\psi_p(\dot\gamma(0)) = \frac{d}{dt}\bigg|_{t=0} (\psi \circ \gamma)(t) = \psi(q)-r $$ so $\phi(q) = \psi(q)$. If $U$ is not convex, $U$ is open, hence a union of convex open sets, on each of which $\phi = \psi$. QED.