Let $V$ be a vector space and $\dim V=n$. Under a basis, the vector $\mathbf v$ is represented by the coordinate $(a_1,a_2,\ldots, a_n)$. Let $S_n$ be the group of all permutations on the set $\{1,2\ldots, n\}$ represented by matrices. $S_n\subseteq M_n(\mathbb R)$.
Let's form the set $$ P_\mathbf v=S_n\mathbf v=\{M\mathbf v:M\in S_n\}. $$
I try to investigate the dimension of $\text{span }P_\mathbf v $. It appears to me that a maximal subset of independent vectors in $P_\mathbf v$ can either be very large ($n$ vectors) or very small (one vector), but not something in between.
What are the possible dimensions of $\text{span }P_\mathbf v $?
The symmetric group of order $n!$ acting as permutation matrices on an n-dimensional vector space (non-modular characteristic) is well-known to have irreducible decomposition given by the trivial representation $1 = \mathrm{span}\{(1, \ldots, 1)\}$ and the $(n-1)$-dimensional "standard representation" $S$. In any case, your $P_v$ is a subrepresentation, so the only possibilities are $P_v \in \{0, 1, S, 1 \oplus S\}$. Correspondingly, the possible dimensions are $0, 1, n-1, n$.
The first two possibilities are realized by $v=0, v=(1, \ldots, 1)$. You can project onto $1$ by applying the averaging ("Reynolds") operator $\frac{1}{n!} \sum_{\sigma \in S_n} \sigma$, which says you get $1$ as a component of $P_v$ if and only if the average of your entries is non-zero. Hence $v=(1, 2, \ldots, n)$ gives you $1 \oplus S$, while $v=(1, -1, 0, \ldots, 0)$ gives you $S$.