 Solution

Use the double covering map $\pi :S^n\to {\mathbb{RP}}^n$ and its deck involution, the antipodal map $a(x)=-x$. The given sphere cohomology vanishes outside degrees $0,n$, so only those degrees can survive as invariant classes. The degree-zero action is the identity because the sphere is connected for $n\ge 1$.

For the top degree, orient $S^n$ as the boundary of the unit ball. A positive tangent basis $(v_1,\dots ,v_n)$ at $x$ means that $(x,v_1,\dots ,v_n)$ has positive determinant in ${\mathbb{R}}^{n+1}$. Under the antipodal map every one of these $n+1$ vectors changes sign, so the orientation sign is $(-1{)}^{n+1}$. Equivalently, the sphere volume form isIts cohomology class is nonzero by the integral argument and therefore generates the one-dimensional top de Rham cohomology. Thus the top class is invariant precisely when $n$ is an odd integer. Applying the quotient result yieldsIn particular ${\mathbb{RP}}^1$ has one copy of $\mathbb{R}$ in each of degrees zero and one. These are real cohomology groups; integral torsion is not detected by de Rham cohomology.