Given a sufficiently nice space $X$, say a connected and compact polyhedron, one has a nice formula for the Poincaré polynomial of the orbit space $SP^n X:= X^n/S_n$ of the $n$-fold Cartesian product under the usual symmetric group action. In particular, writing $b_i$ for the $i$th Betti number of $X$:
$ p_t(SP^nX) = \text{the } x^n \text{ coefficient of } \frac{(1+tx)^{b_1}(1+t^3x)^{b_3} \cdots}{(1-x)(1-t^2x)^{b_2}\cdots}. $
One can then compute that, for $X = S^{2k+1}$ an odd-dimensional sphere, we have
$p_t(SP^n S^{2k+1}) = 1 + t^{2k+1} = p_t(S^{2k+1}).$
In fact, when $k=0$, this equality comes from a homotopy equivalence. There is always the inclusion
$i: X \hookrightarrow SP^nX \qquad i(x) = x + x_2 + \cdots + x_n,$
where the $x_i$ are fixed constants and we use additive notation for elements of $SP^n X$. On the other hand, thinking of $S^1$ as the unit complex numbers, we can write down a retract
$r: SP^nS^1 \to S^1 \qquad r( \, \sum_i x_i \, ) = \prod_i x_i.$
One can then show that $r$ and $i$ together constitute a homotopy equivalence $S^1 \simeq SP^n S^1$.
Now this is all wildly false for even-dimensional spheres. In particular, one can show $SP^n S^2 \cong \mathbb{CP}^n$. Also the previously mentioned homotopy equivalence does not hold for odd spheres: in light of the cofibration (see these notes if interested)
$ \Sigma^n \mathbb{RP}^{n-1} \to S^n \hookrightarrow SP^2 S^n $
we have the formula $H_*(SP^2(S^n)) \cong H_*(S^n) \oplus \tilde{H}_*(\Sigma^n \mathbb{RP}^{n-1})$, so there is 2-torsion in $SP^2(S^{2k+1})$ whenever $k>0$. However I'm curious if one can construct a map
$SP^n S^{2k+1} \to S^{2k+1}$
inducing an isomorphism on $H_*(-;\mathbb{Q})$. The only idea I've got, thinking of $S^{2k+1} \subset \mathbb{C}^{k+1}$, is
$f: SP^n S^{2k+1} \to \mathbb{C}^{n(k+1)}-0 \qquad f((b^1_0,\dots,b^1_k)+\cdots+(b^n_0,\dots,b^n_k)) = (\dots, e_i(b_j), \dots),$
where $e_i$ is the $i$th elementary polynomial in $n$ variables and $b_j$ stands for the tuple $(b^1_j,\dots,b^n_j)$, defined up to permutation. This mimics the usual formulas used to show $SP^n \mathbb{C}^m \cong \mathbb{C}^{nm}$.
Of course, in this attempt, the dimension of the image is much too large. The big trouble with directly generalizing to $k>0$ seems to be that higher-dimensional spheres no longer have an abelian group structure, and I can't think of an alternative way to take an $n$-tuple of vectors on the sphere and produce a new non-zero vector in a symmetric way.