= Solution
Under a smooth coordinate change $x\mapsto x'$, a <contravariant vector> obeys
$$
\boxed{T'^a=\frac{\partial x'^a}{\partial x^b}T^b.}
$$
The <scalar>-contraction hypothesis gives $U'_a(\partial x'^a/\partial x^b)T^b=U_bT^b$ for every choice of $T^b$. Equality of all coefficients and multiplication by the inverse <Jacobian matrix> give $\boxed{U'_a=(\partial x^b/\partial x'^a)U_b}$, the <covector> law.
Let $V^a=dx^a/ds$ be a <timelike geodesic> tangent. <Metric compatibility> and the <geodesic equation> give
$$
\frac d{ds}(T_aV^a)=V^aV^b\nabla_bT_a+T_a\frac{DV^a}{Ds}
=V^aV^b\nabla_{(a}T_{b)}.
$$
A <Killing vector> therefore gives a conserved contraction. Conversely, if that contraction is conserved along every <timelike geodesic>, then at any point its symmetric <tensor> $S_{ab}=\nabla_{(a}T_{b)}$ satisfies $S_{ab}V^aV^b=0$ for every unit timelike initial tangent. Homogeneity extends this to the entire <timelike cone>, an open set of tangent vectors. A <quadratic polynomial> vanishing on an open set has every coefficient zero, so $\boxed{\nabla_{(a}T_{b)}=0}$. Considering a single selected <geodesic> would not suffice. This is the <Converse geodesic characterization of a Killing vector>.
For the given exterior transformation, differentiating at fixed $r$ gives
$$
\boxed{T=\partial_t=-\frac{U}{4m}\partial_U+\frac{V}{4m}\partial_V.}
$$
The transformation also gives $UV=(2m-r)e^{r/(2m)}$. This determines $r$ smoothly for $r>0$ throughout $UV<2m$. The flow of the displayed vector field rescales $U\mapsto e^{-\lambda/(4m)}U$, $V\mapsto e^{\lambda/(4m)}V$. It leaves $UV$, $r$, $dU\,dV$ and the angular metric unchanged. Thus it preserves the full metric everywhere in that region and extends as a global <Killing field>, not merely by a formal exterior coordinate substitution.
Its squared norm is
$$
\boxed{g(T,T)=-\frac{UV}{r}e^{-r/(2m)}=1-\frac{2m}{r}.}
$$
It is \b[timelike] for $UV<0$ ($r>2m$), \b[spacelike] for $0<UV<2m$ ($0<r<2m$), and a nonzero \b[<null vector>] on either horizon $UV=0$ away from $U=V=0$. At $U=V=0$ it is the \b[<zero vector>], on the whole <bifurcation surface> of angular points. Zero must be distinguished from a nonzero <null vector>. This is the <Schwarzschild time-translation Killing field in Kruskal coordinates>.
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