 Solution

Let $a(x)=-x$ be the antipodal map, whose degree is $(-1{)}^{n+1}$. The invariance condition is $f\circ a=f$. Multiplicativity of degree gives $\mathrm{deg}(f)(-1{)}^{n+1}=\mathrm{deg}(f)$, so if $n$ is even,For odd $n$, factor the invariant map uniquely through the quotient $q:S^n\to {\mathbb{RP}}^n$, writing $f=gq$. The standard orientability and homology facts for Real projective space are that ${\mathbb{RP}}^n$ is orientable in odd dimension and has top integral homology $\mathbb{Z}$. With compatible orientation the double cover $q$ has degree two: both local sheets preserve orientation because the antipodal map does. Thus $g_{\ast }$ is multiplication by an integer $k$ and $\mathrm{deg}(f)=2k$.

Every such integer occurs. In the oriented projective space, choose a small oriented closed ball and collapse its exterior to a point. The resulting map $c:{\mathbb{RP}}^n\to S^n$ has degree one, because a point in the interior image has one orientation-preserving preimage. Composing it with a degree-$k$ sphere map from part (b) givesHere ${\alpha }_k$ denotes the dimension-$n$ map constructed by suspension. The factorization through $q$ makes the resulting map antipodally invariant.