It seems to me that your question has not so much to do with Killing fields. It is a more general question. Consider a smooth vector field $X$ over a smooth (Hausdorff) manifold $M$ and suppose that the one-parameter group of local diffeomorphisms $\phi$ associated to $X$ is global (which is equivalent to saying that $X$ is complete). In other words, if $x\in M$ the differential equation $$\dot{\gamma}_x(t) = X(\gamma_x(t))$$ with initial condition $$\gamma_x(0)=x$$ admits a (unique) maximal solution $\gamma_x= \gamma_x(t)$ defined for all $t \in \mathbb R$.
There are sufficient conditions assuring that $\phi$ is global (for instance it happens provided $M$ is compact).
This way, $\phi : \mathbb R \times M \ni (t,x) \mapsto \phi_t(x):= \gamma_x(t) \in M$ is smooth and well-defined. Moreover
(1) $\phi_0 = id$
and
(2) $\phi_t \circ \phi_\tau = \phi_{t+\tau}$ for every $t,\tau \in \mathbb R$.
The case you are considering also requires that $M$ is equipped with a nondegenerate metric $g$ and $X$ is a complete $g$-Killing vector field.
In this case every $\phi_t : M \to M$ is an isometry.
Well, coming back to the general case, the following proposition is valid.
PROPOSITION. Let $A \subset M$ be an open set whose boundary $\partial A$ is a smooth codimension-$1$ embedded submanifold of the smooth manifold $M$ and $X$ a smooth complete vector field on $M$.
Then the following two facts are equivalent.
(a) $\phi_t(A) = A$ and $\phi_t(M\setminus \overline{A}) = M\setminus \overline{A}$ for every $t \in \mathbb R$.
(b) $X$ is tangent to $\partial A$.
Proof.
(1) We prove that not (a) implies not (b).
If it is false that $\phi_t(A) = A$ and $\phi_t(M\setminus \overline{A}) = M\setminus \overline{A}$ for all $t$, then there must exist a point $x_0 \in A$ such that $\phi_{t_0}(x_0) \not \in A$ or a point $x_0 \in M\setminus \overline{A}$ such that $\phi_{t_0}(x_0) \not\in M \setminus \overline{A}$ for some $t_0 \in \mathbb R$. Assume the former is valid (the latter can be treated similarly). Assume $t_0>0$ the other case is analogous.
There are now two possibilities for $\phi_{t_0}(x_0) \not \in A$. One is $\phi_{t_0}(x_0)\in \partial A$ and in this case define $s:= t_0$. The other possibility is $\phi_{t_0}(x_0)\in M \setminus\overline{A}$. In this case, define
$$s := \sup\{t \in [0,+\infty) \:|\: \phi_\tau(x_0) \in A\:, \quad \tau < t\}\:.$$ This number exists and is finite (because the set is not empty, as it contains $0$, and $t_0<+\infty$ is an upper bound), strictly positive not greater than $t_0$, and again $\phi_s(x_0) \in \partial A$.
(Indeed, if $\phi_s(x_0)\in A$ there is an open neighborhood of $\phi_s(x_0)$ completely included in $A$ so that $\phi_\tau(x_0) \in A$ also for some $\tau>s$ which is impossible for the very definition of $\sup$, if $\phi_s(x_0)\in M\setminus \overline{A}$, since this set is open, there would be an open neightborhood of $\phi_s(x_0)$ completely included in $ M\setminus \overline{A}$ so that $\phi_\tau(x_0) \not\in A$ in some $(s-\epsilon, s]$ which is again impossible for the very definition of $\sup$; the only remaining case id $\phi_s(x_0) \in \partial A$.)
Let us prove that such $s$ (in both possibilities) cannot exist if (b) is valid. Indeed, $X|_{\partial A}$ is a well-defined smooth complete vector field on the smooth manifold $\partial A$ and thus the associated Cauchy problem over $\partial A$ with initial condition $\dot{\gamma}(s)= \phi_s(x_0) \in \partial A$ at $t=s$ admits a complete solution completely contained in $\partial A$ also for $t<s$, but this curve now viewed as a integral line of $X$ in $M$ is uniquely determined and we know by hypothesis that it starts at $x_0 \not \in \partial A$ finding a contradiction.
(2) We prove that not (b) implies not (a).
Let us assume that (b) is false finding that (a) is false as well.
Assume now that there is $x_0 \in \partial A$ such that $X(x_0)$ is transverse to $\partial A$. As $\partial A$ is an embedded smooth manifold, $X$ is smooth and does not vanish at $x_0$, it is not to difficult to prove that there is a coordinate patch $x^1,x^2,..., x^n$ around $x_0$ in $M$ ($n = dim(M)$) such that $x_0 \equiv (0,0,\ldots, 0)$,
$\partial A$ is the portion of the plane $x^1=0$ contained in the image of the chart, and
the integral
curves of $X$ are the curves $\mathbb R \ni t \mapsto (t,x^2,\ldots,x^n)$ (see the final ADDENDUM). Since the plane separates $A$ from $M \setminus \overline{A}$, it is evident that there are points in $A$ which are moved into $M \setminus \overline{A}$ by $\phi$ and viceversa. Therefore $\phi_t(A) = A$ and $\phi_t(M\setminus \overline{A}) = M\setminus \overline{A}$ for every $t \in \mathbb R$ is false.
QED
Evidently, if $X$ is a complete Killing field, the result concerns the associated one-parameter group of isometries.
ADDENDUM. I prove here that
Lemma. If $S$ is an embedded $n-1$-dimensional smooth submanifold of the $n$-dimensional smooth manifold $M$, and $X$ is a smooth vector field over $M$ which does not vanish at $x_0\in S$ and is not tangent (i.e., is transverse) to $S$ at $x_0$, then there is a coordinate patch $x^1,x^2,..., x^n$ around $x_0$ in $M$ such that $x_0 \equiv (0,0,\ldots, 0)$,
$S$ is the portion of the plane $x^1=0$ contained in the image of the chart, and the integral curves of $X$ are (restrictions around $t=0$ of) the curves $\mathbb R \ni t \mapsto (t,x^2,\ldots,x^n)$.
Proof. As $S$ is embedded, there is a coordinate patch $(U, \psi)$ in $M$ around $x_0\in S$ such that $\psi(S \cap U) = \{ (y^1,\ldots, y^n) \in \psi(U) \:|\: y^1=0\}$ and we can always assume $\psi(x_0)= (0,\ldots,0)$.
Now $X = \sum_a Y^a\frac{\partial}{\partial y^a}$ is such that $Y^1(0,\ldots, 0) \neq 0$ just because $X$ is transverse to $S$ at $x_0$ (the coordinates $y^2,\ldots, y^n$ are coordinates on $S$). The integral lines of $X$ in coordinates satisfy $\frac{dy^a}{dt} = Y^a(y^1(t),\ldots, y^n(t))$. We are free to fix $t=0$ exactly on $S$ for all curves. Now introduce the coordinates $x^2 =y^2,\ldots, x^n=y^n$ on $S$ and write the said integral curves as smooth functions $y^k = y^k(t,x^2,\ldots, x^n)$, where $x^2,\ldots, x^n$ denotes the initial point on $S$ (at $t=0$) of the considered integral curve. The said map is smooth as well known from standard theorems on smooth dependence from initial data of Cauchy problems.
Finally define $x^1=t$. Since the Jacobian matrix $J=[\frac{\partial y^a}{\partial x^b}]$ exactly at $x_0$ satisfies $$\det J(x_0) = \frac{\partial y^1}{\partial t}|_{x_0} = Y^1(0,\ldots, 0) \neq 0$$
Dini's theorem proves that $x^1=t$, $x^b= y^b$ define an admissible smooth coordinate system in $M$ around $x_0$. In local coordinates $x^1,\ldots, x^n$, the portion of $S$ entering the domain of the coordinates is still represented by $x^1=0$ (because $x^1=t$ and all integral curves intersect $S$ at $t=0$) and, locally, the integral curves of $X$ are trivially restrictions around $t=0$ of the curves $\mathbb R \ni t \mapsto (t,x^2,\ldots,x^n)$.
