Consider each linear function as a column vector in the dual space with some basis. Then the tensor product of $n$ such vectors can be visualized as a hyper matrix (or tensor, as it's usually described in physics) of dimension $n$, where each entry is the product of entries in each column vector. Two vectors $a,b$ tensored gives a matrix where each row is a multiple of the first vector, and the multiple of each row is given by the entries in the second column vector.
Tensoring such a matrix with a third tensor gives a cubical matrix where each horizontal slice is a multiple of the base matrix, and the multiples are given by the third column vector, etc.
Note that every single row (i.e. 1-dimensional subset) in the $i$th direction of the tensor hyper matrix is a multiple of the $i$th vector in the tensor product. Either every such row is zero (and the tensor is zero) or we can recover the $i$th vector in the tensor product up to a scalar factor.
This means that in non-zero tensor products, the multiplicands and their order are uniquely determined up to scalar products. Switching the order of tensoring as described in the problem is equivalent to taking an $r+s$ dimensional tensor permuting the order of the multiplicands by a cyclic permutation (I.e. shifting every index by $s$ and modding indices by $r+s$). If the tensor is invariant under such a cyclic permutation, then the multiplicands of the product tensor must be, up to a scalar factor,periodic with period $gcd(s,r+s)=gcd(s,r)$, i.e. the 2 vectors being tensored must be scalar multiples of tensor powers of the same tensor.