Let $E$ be a finite-dimensional Euclidean space and $w$ an isometry of $E$ (which is necessarily invertible and affine-linear by the Mazur-Ulam theorem). Let $S$ be a subset of $E$ such that
$$w(S)\subseteq S$$.
Is it then necessarily true that $w(S)=S$? Or equivalently, that $w^{-1}$ preserves $S$ as well?
Intuitively, it feels like it must be true, but I can't find a source or think of a proof. I'm adding algebraic geometry as a tag in case there's some jackhammer there that can be applied.
No. Think about the right shift of the positive integers on the real line.
In a positive direction, the statement is true if $S$ is compact. In this case there is a nice general argument that every isometry of a compact metric space is bijective.
The best we could hope for is that the statement is true if $S$ is measurable with finite volume, as Ethan suggests in the comments; I don't know if it's true or not in this level of generality. The pieces of the Banach-Tarski decomposition are a counterexample even for $w$ a rotation if we don't require measurability. And on the other hand if $S$ has finite volume then it has the same volume as $w(S)$, so $S \setminus w(S)$ has measure zero.
Since there are trivial counterexamples from translations, let's restrict to the case that $w$ fixes some point, which might as well be declared the origin. In other words, let's work in $\mathbb{R}^n$ and fix $w \in O(n).$
This isn't enough to prevent counterexamples. For example, let $n=2$ and $w$ be rotation by 1 radian. Then $S= \{(\cos n, \sin n)\}_{n \in \mathbb{N}}$ and $T = S^1 \setminus \{(\cos n, -\sin n)\}$ each have the property $w(S) \subsetneq S.$ Both are measure 0 in the plane, but since $T$ is a full measure subset of the unit circle, we can get counterexamples of arbitrary plane measure by considering sets of the form $\bigcup_{\lambda \in [0, c]} \lambda T.$
However, all Borel sets satisfying $w(S) \subset S$ admit such a decomposition into measure 0 and 1 layers. First we'll show that for $p, q \in \mathbb{R}^n,$ the relation $p \sim q$ defined by $``q$ is in closure of $\{w^n(p)\}_{n \in \mathbb{N}}"$ is an equivalence relation, and each class $[p]$ is a disjoint union of finitely many isometric copies of some $k$-torus $\prod_{i=1}^k \lambda_i S^1,$ some $k \in [0, \lfloor n/2 \rfloor].$ Furthermore, $w$ acts ergodically on each equivalence class.
Assume that $w \in SO(n)$ (the other case follows from considering $w^2$). Choose an orthonormal basis of $\mathbb{R}^n$ for which $w$ is in canonical matrix form, so that $w = R_{\theta_1} \oplus \ldots \oplus R_{\theta_j} \oplus I_{n-2j}$ (where $R_{\theta}$ is 2D rotation by $\theta$), and that $p$ has infinite orbit under $\langle w \rangle.$ Let $\lambda_i'=\|(p_{2i-1}, p_{2i})\|,$ which we may assume to be decreasing in $i$ by reordering the basis. Let $j'$ be greatest such that $\lambda'_{j'}>0.$ Then $w$ acts on the $j'$-torus $T=\prod_{i \le j'} \lambda'_i S^1.$ By Kronecker-Weyl, the closure $C$ of $\{w^n(p): n \in \mathbb{N}\}$ is a disjoint union of finitely many isometric $k$-tori, where $k=-1+\dim_{\mathbb{Q}}\mathrm{span}_{\mathbb{Q}}(1 \cup \{\theta_i/(2\pi): i \le j' \}),$ and that $w$ acts ergodically on $C.$
For $S$ Borel with $w(S) \subset S$ and $M = [p]$ for some $p \in \mathbb{R}^n,$ $S \cap M$ is a Borel set with 0 or full measure, relative to Lebesgue measure on $M$ as a submanifold (if we merely assume $S$ is measurable, we must weaken this to almost all $p$ are such that $S \cap [p]$ has 0 or full measure relative to $[p].$) This follows from ergodicity of $w$ on $M$ and the fact that $\mu_M(M \cap S \setminus w(S))=0.$
(If we merely assume $S$ is measurable, we must weaken this to almost all $p$ are such that $S \cap [p]$ has 0 or full measure relative to $[p].$)
Finally, here's a fairly general condition which (in addition to $w(S) \subset S$) guarantees that $w(S)=S.$ Suppose $S$ is $\Delta_2,$ i.e. $S$ and $\mathbb{R}^n \setminus S$ are both $F_{\sigma}$ sets. Fix $p \in S$ and let $M=[p].$ We will show $w^{-1}(p) \in M \subset S.$ Suppose not. Fix $q \in M \setminus S.$ We have $M \cap S = \bigcup_{i \in \mathbb{N}} C_i$ and $M \setminus S = \bigcup_{i \in \mathbb{N}} D_i,$ some closed $C_i, D_i.$ Each $D_i$ is nowhere dense in $M$ since $M \setminus D_i$ contains the dense-in-$M$ set $\{w^n(p)\}_{n \in \mathbb{N}},$ and similarly $C_i$ is nowhere dense in $M$ since $M \setminus C_i$ contains $\{w^{-n}(q)\}.$ This contradicts the Baire category theorem for $M.$ Thus, $p \in w(S).$
