Let $V$ be the subspace spanned by the permuted versions of the vector $a=(a_1,a_2, ... ,a_n) \in K^n$. Let $U$ be the complementary subspace w.r.t. the usual bilinear form
$\langle x,y\rangle=\sum_{i=1}^nx_iy_i$ of $K^n$, i.e.
$$
U=\{ x\in K^n\mid \langle x,y\rangle=0\;\text{for all $y\in V$}\}
$$
Then it is impossible for both $V$ and $U$ to have a vector with non-equal components.
Assume contrariwise that $a_i\neq a_j$ for some pair of indices $i
The conclusion is that one of the subspaces $U$ or $V$ is of dimension at most 1. As they are complementary $\dim U +\dim V=n$, and we can deduce that the dimension of $V$ is either 0,1, $n-1$ or $n$.
It is clear, when either of the first two cases occurs. If $\dim V=n-1$, then $\dim U=1$,
and, by the above observation, $U$ is spanned by the all ones vector, so $V$ is the zero-sum subspace. When none of these cases applies, we must have $\dim V=n$.