 Solution

The antipodal map is the restriction of the smooth linear map $-I$, so is smooth. For $n\ge 1$ orient the sphere by the outward-normal convention. Its degree is $(-1{)}^{n+1}$: the ambient determinant is $(-1{)}^{n+1}$, and the outward normal is carried to the outward normal at the antipodal point.

Choose a regular value of $f$, available by Sard theorem; if it is outside the image, the degree is zero already. Otherwise its finite preimage is partitioned into distinct antipodal pairs. Differentiating $f\circ h=f$ shows that their local degree signs obeyFor even $n$, the signs in each pair cancel, so $\mathrm{deg}f=0$. For odd $n$, each pair contributes $2$ or $-2$, so $\mathrm{deg}f$ is even. In neither case can its degree equal the identity's degree one. Since the degree of a map between oriented manifolds is invariant under smooth homotopy, $f$ is not smoothly homotopic to the identity. For $n=0$, the symmetry makes $f$ constant on the two-point discrete sphere; a continuous homotopy into this discrete space cannot change either point's image, so the conclusion still holds.