= Solution
The <antipodal map> is the restriction of the smooth linear map $-I$, so is smooth. For $n\geq1$ 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 obey
$$
\varepsilon_f(-x)(-1)^{n+1}=\varepsilon_f(x).
$$
For even $n$, the signs in each pair cancel, so $\deg f=0$. For odd $n$, each pair contributes $2$ or $-2$, so $\deg f$ 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>, \b[$f$ 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.
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