Which groups can be reconstructed from a single invariant subspace? Let $G\subseteq\mathrm{Perm}(\Bbb R^n)$ be a matrix group consisting of permutation matrices acting on $\Bbb R^n$. Let $U\subseteq\Bbb R^n$ be an irreducible invariant subspace w.r.t. $G$. Now, define the reconstructed group
$$G_U:=\{T\in\mathrm{Perm}(\Bbb R^n) \mid TU=U\},$$
of permutation matrices that leave $U$ invariant. Obviously $G\subseteq G_U$.

Question: For which groups $G\subseteq\mathrm{Perm}(\Bbb R^n)$ there is some irreducible invariant subspace $U\subseteq\Bbb R^n$ that already determines $G$, i.e. $G=G_U?$


Some notes


*

*This problem is related to reconstructing such a group from all its invariant subspaces, and this is related to finding GRRs (graphical regular representations). E.g. a cyclic group cannot be reconstructed from any invariant subspace, as a suitable dihedral group has the exact same invariant subspaces, but contains more permutation matrices.

*So assuming that $G$ can be reconstructed from knowing all invariant subspaces $U_1,...,U_k$, i.e. $$G=\{T\in\mathrm{Perm}(\Bbb R^n)\mid TU_i=U_i,i=1,...,k\},$$ when can we find a single subspace $U_i$ that suffices for reconstruction? Actually, I do not know a counterexample where we cannot find such a subspace.

*I am especially interested in the case wher $G$ acts regularly (hence transitively) on the canonical base vectors of $\Bbb R^n$.
 A: Here is a counterexample like you asked for: $G=V_4=\{1,(12)(34),(13)(24),(14)(23)\}$. The irreducible subspaces are one-dimensional and spanned by $(1,1,1,1)^t$, $(1,-1,1,-1)^t$, $(1,1,-1,-1)^t$, and $(1,-1,1,-1)^t$ respectively. The permutation groups stabilising these subspaces are $Sym(4)$ and the three 2-sylowgroups of $Sym(4)$. Their intersection is $V_4$ so that $G$ is reconstructible from all of its irreducible subspaces combined, but not any single one individually. This example also happens to be a regular action.

(The following is based on the ideas of Rebecca Waldecker, Paula Hähndel, ... on orbital graphs)
Now some thoughts on your general question, even though not a full answer.
Let $\Omega$ be a finite $G$-set and $V:=\mathbb{C}^\Omega$ the associated $\mathbb{C}G$-permutation module.
First observe that $Perm(n)\leq U(n)$ and every $G$-invariant subspace corresponds to a unique self-adjoint idempotent $p_U$ (the orthogonal projection onto $U$) in $End_{\mathbb{C}G}(V)$ and vice versa. Standard results from representation theory tell us that this endomorphism ring is isomorphic to a direct sum of matrix rings. And for complex matrix rings it is an easy exercise to show that there are enough self-adjoint idempotents in them to span the whole algebra. (Here is where I use the special properties of $\mathbb{C}$ for the first time. For $\mathbb{R}^{n\times n}$ this is not true, because the space of symmetric matrices is too small.)
What does that mean in your situation? Being in $G_U$ means commuting with $p_U$. Being in $\bigcap_{U\leq V\;G\text{-inv.}} G_U$ therefore means commuting with all the idempotents $p_U$ and therefore with all of $End_{\mathbb{C}G}(V)$ so that $g\in\bigcap G_U$ iff $g$ is a permutation matrix and in the centraliser of $End_{\mathbb{C}G}(V)$. This proves

Lemma: A permutation group $G\leq Perm(n)$ is reconstructible from all its invariant subspaces combined if and only if $G = Perm(n)\cap C(End_{\mathbb{C}G}(V))$.

More generally we define:

Definition: $G\leq Perm(n)$ is reconstructible from $\mathcal{X}\subseteq End_{\mathbb{K}G}(V)$ if $G=Perm(n) \cap C(\mathcal{X})$.

This group $Perm(n) \cap C(End_{\mathbb{K}G}(V))$ is also called the 2-closure of $G$, i.e. it is the largest subgroup of $Sym(\Omega)$ that has the same orbits on $\Omega^2$ as $G$ (those are called orbital). Explicitly writing down what it means for a matrix to be in $End_{\mathbb{C}G}(V)$ shows that $End_{\mathbb{C}G}(V)$ has a basis $\{X_\Gamma\}$ of the form
$$(X_\Gamma)_{ij} := \begin{cases} 1 & (i,j)\in\Gamma \\ 0 & \text{otherwise} \end{cases}$$
where $\Gamma=\Gamma_1,\ldots,\Gamma_r$ runs through the orbitals of $G$. Therefore the 2-closure of $G$ has the same endomorphism ring and therefore the same centraliser of its endomorphism ring.

Corollary: A permutation group is reconstructible from all of its invariant subspaces iff it is equal to its 2-closure iff it is reconstructible from $X_{\Gamma_1},\ldots,X_{\Gamma_r}$.

This corollary is also useful from a computational perspective: There might be infinitely many invariant subspaces (namely iff some irreducible constituent of the $\mathbb{C}G$-module $V$ occurs with multiplicity $>1$) and a priori it is not clear how many one needs to find the 2-closure. But there are always only finitely many orbits on $\Omega^2$ so that the 2-closure is computable in a set amount of time. Additionally: Finding the invariant subspaces in the first place is harder than the purely combinatorial exercise of finding the orbitals.
It is useful to have a closer look at the other direction: If $G$ is reconstructible from some matrix $X$ (or some set of matrices), what is some explicit set of invariant subspaces from which we can reconstruct $G$ ?

Lemma: A group $G\leq Perm(n)$ is reconstructible from $X_1,\ldots,X_r\in End_{\mathbb{C}G}(V)$ iff it is reconstructible from the collection of subspaces
  $$\left\{Eig_\lambda(\mathfrak{Re}(X_s)), Eig_\lambda(\mathfrak{Im}(X_s)) \mid \lambda\in\mathbb{R}, s\in\{1,\ldots,r\}\right\}$$

Proof: Every matrix can be written as $X=\mathfrak{Re}(X) + i\mathfrak{Im}(X)$ as usual by defining $\mathfrak{Re}(X) = \tfrac{1}{2}(X+X^\ast), \mathfrak{Im}(X):=\tfrac{1}{2i}(X-X^\ast)$. Hence $g\in U(n)$ centralises $X$ iff it centralises both real and imaginary part.
Real and imaginary part are self-adjoint matrices. The spectral theorem shows that every self-adjoint $Y$ is a complex (even real) linear combination of its eigenspace projections. Lagrange interpolation shows that every eigenspace projection is a polynomial in $Y$. Hence $g\in U(n)$ centralises $Y$ iff it centralises the eigenspace projections. QED.
How does this help us? If we can find a group that is reconstructible from the collection of all its orbitals (i.e. it is 2-closed), but not a single one, then we have found a group that is reconstructible from the set of eigenspaces in the list above, but not from a subspace on this list, because centralising a single eigenspace $Eig_\lambda(Y_s)$ for one $\lambda\in\mathbb{R}$ and one $s\in\{1,...,r\}$ is not enough, even centralising all eigenspaces of $Y_s$ wouldn't be enough.
Some easy consequences:


*

*If $G$ is 2-closed and 2-transitive, then $G=Sym(n)$ is reconstructible from each of its invariant subspaces.

*If $G$ is 2-closed and has rank 3, then the two non-diagonal orbitals are complementary graphs and therefore have identical automorphism groups (not isomorphic, identical). Therefore $G$ is reconstructible from each of its non-diagonal orbitals. If they happen to be paired to each other (i.e. if $\Gamma_1^{op} = \Gamma_2$), then the symmetric part $X_\Gamma + X_\Gamma^t$ has no useful eigenspaces, i.e. only $\mathbb{C}(1,\ldots,1)$ and its orthogonal complement. The skew-symmetric part on the other hand has at least one pair of eigenvalues of the form $\pm i r$ (because it is non-zero) and also it has $(1,\ldots,1)^t$ in its kernel. Its minimal polynomial has degree $\leq \dim End_{\mathbb{C}G}(V)=rank=3$ so that the minimal polynomial must have the form $T(T+r^2)$ for a single $r\in\mathbb{R}$. Thus there are exactly three eigenvalues and the three eigenspaces must be equal to the three irreducible components of $V$. (Note that this also does not work over $\mathbb{R}$. There are only the two obvious irreducible subspaces over $\mathbb{R}$)
Consequently $G$ is reconstructible from each of its irreducible components.

*If $G$ acts regular, then if we identify $\Omega$ with $G$ endowed with left multiplication, the orbitals are $\Gamma_h=\{(x,y) \mid x^{-1}y=h\}$. Every permutation stabilising all those orbitals is easily seen to be a left multiplication itself. Therefore $G$ is automatically 2-closed. The matrices $X_{\Gamma_h}$ are exactly the permutation matrices coming from the right-multiplication action on $G$.

*In the example $G=V_4=\langle (12)(23),(13)(24)\rangle$, the first non-diagonal orbital is $\Gamma_1:={^G(1,2)}=\{(1,2),(2,1),(3,4),(4,3)\}$. The other two orbitals are $Sym(4)$-conjugated. The automorphism groups of the orbital graphs $(\{1,2,3,4\},\Gamma_i)\}$ are exactly the 2-Sylowgroups of $Sym(4)$ so that neither of them is sufficient to reconstruct $V_4$, but all of them together are. The graphs are undirected so that $X_\Gamma$ is symmetric. The eigenspaces of $X_\Gamma$ for the eigenvalues 1 contain $(1,1,1,1)^t$ of course, but they are two-dimensional so that we get another one-dimensional invariant subspace in each case. These are the invariant subspaces I gave at the top of the post.
