The answer is "it depends". Consider, for example, the permutations of (1,1,1,...1) and the permutations of (1,0,0,...,0).
Well, that seems to be the case. The proof is really straightforward, let have a nonzero entry a at the i-th place: . We can take , as otherwise we can consider .
Let be the subspace created by and its permutations, and consider the j-axis of .
We consider a permutation of with 1 occupying the j-th place, and denote the vector obtained by .
where is the jth euclidean basis vector. And
This proves contains the j-th axis of . So contains every axis, and therefore .
In working up a proof of my conjecture, I found a counter example and had to revise the conjecture. (Rebesques, take note. :( ) Here is the revised conjecture and a proof.
Let The dimension of the subspace spanned by and its permutations is
As examples in , for while for
Claims C and D are proven by showing that the given number is the maximum number of independent linear combinations of and its permutations. So assume for any Under this assumption we may further assume wlog that
since we can permute as we like. Let
We will use the following linear combinations of and its permutations.
Putting these vectors into a matrix
it is clear that the first vectors are linearly independent and their elements sum to zero. If the elements of the last vector do not sum to zero, it will be independent of the first vectors and the number of independent vectors in will be the maximum possible. This proves claim D.
Proving C, if the elements of the last vector sum to zero, it will be linearly dependent on the first vectors. Since the last vector could represent any permutation of , this means there cannot be more than independent vectors in This proves C and completes the proof.