Take a singular $n$-simplex $\sigma: \Delta^n \to X$, where $\Delta^n\subset \mathbb{R}^{n+1}$ is the convex hull of the standard basis, with the obvious vertex ordering. Then one can obtain $(n+1)!$ singular simplices from this by permuting the standard basis via linear maps and composing with $\sigma$. My intuition is that two such singular simplices will only differ up to some boundary of a singular $(n+1)$-chain (additively) and up to the sign of the permutation (multiplicatively), since homology should only depend on the orientation of a simplex. More precisely, if $s$ is a permutation as described above and $\sigma_s$ is the simplex thus obtained, I expected $$\sigma = (-1)^s \sigma_s +\partial c$$ for some $n+1$-chain $c$, with the image of $c$ contained in the image of $\sigma$.
Edit: As it was pointed out to me, this exact relation doesn't hold. The statement that I ended up using was as follows. Consider the $n$-simplex $\sigma$ in $\mathbb R^n$ given as the convex hull of the standard basis and the vector $(-1, \dots, -1)$, interpreted as a singular simplex. Again we can obtain $n!$ singular simplices $\sigma_s$ from this by permuting the standard basis with permutations $s$. One can show that $\sigma_s = (-1)^s \sigma \in \mathrm H^n(\mathbb R^n, \mathbb R^n\backslash \{0\}).$