$ S_n $ in $ SO_{n-1}(\mathbb{R}) $ and the adjoint representation $\operatorname{S}_n$ has a well known real representation of degree $n - 1$. In other words, $\operatorname{S}_n$ is a subgroup of $\operatorname{SO}_{n-1}(\mathbb{R})$.
Recall, the adjoint representation is reducible for $\mathfrak{so}_{4}(\mathbb{R})$.
Is it true that this $\operatorname{S}_n$ subgroup acts irreducibly on $\mathfrak{so}_{n-1}(\mathbb{R})$ in the adjoint representation for all $n \neq 5?
I know that $\operatorname{S}_4$ is irreducible in the natural rep of $ \operatorname{SO}_3(\mathbb{R})$ which also happens to be the adjoint rep.
For higher $\operatorname{S}_n,\: n \geq 6$ one can check character tables, and it seems they all have an irrep of $\operatorname{dim}(\mathfrak{so}_{n-1}(\mathbb{R}))$ and that this irrep always has Frobenius-Scur indicator 1.
But this is just circumstantial evidence. Is there a direct way to show that the $\operatorname{S}_n$ subgroup of $\operatorname{SO}_{n-1}(\mathbb{R})$ acts irreducibly in the adjoint representation on $\mathfrak{so}_{n-1}(\mathbb{R})$?
 A: If you take the standard $ n-1 $ dimensional deleted permutation representation $ V $ of $ S_n $ then the answer on Is the alternating group $ A_8 $ a subgroup of the exceptional lie group $G_2$? claims that all of its exterior powers $ \bigwedge^k(V) $ are irreducible, combinatorially they correspond to so-called hook shaped partitions $ (n-k,1^k) $. For $ A_n $ the story is similar with the slight exception that if $ n=2k+1 $ is odd then $ \bigwedge^k(V) $ decomposes into two parts.
Furthermore, the answer to Decomposition of $ V \otimes V^* $ for the natural representation cites Fulton and Harris that $ \bigwedge^2 $ is the adjoint representation for the orthogonal group. So
$
 \bigwedge^{2}(V) \cong \mathfrak{so}_{n-1}(\mathbb{R}) 
$
where again $ V $ is the $ n-1 $ dimensional deleted permutation irrep. So for $ n \neq 5 $ then $ \mathfrak{so}_{n-1}(\mathbb{R}) $ is irreducible as both an $ S_n $ and an $ A_n $ irrep.
For the exceptional case
$$
 n=5=2(2)+1=2k+1 
$$ then the adjoint rep of $ SO_4(\mathbb{R}) $ on  $ \mathfrak{so}_{4}(\mathbb{R}) $ is reducible so the adjoint rep restricted to any $ A_5 $ subgroup of $ SO_4(\mathbb{R}) $ is also reducible.
However the adjoint rep of $ O_4(\mathbb{R}) $ on  $ \mathfrak{o}_{4}(\mathbb{R})=\mathfrak{so}_{4}(\mathbb{R}) $ does not reduce. And it turns out that, as expected, any $ S_5 $ subgroup of $ O_4(\mathbb{R}) $ lies outside of $ SO_4(\mathbb{R}) $ and acts irreducibly in the adjoint representation.
Also thanks to @MarianoSuárez-Álvarez !
