Homotopy classes of maps from projective plane to projective plane

There are at least $\Bbb Z$-many unbased homotopy classes of maps.

First, an odd map $S^2 \to S^2$ (that is, one such that $f(-x) = -f(x)$) descends to a map $\Bbb{RP}^2 \to \Bbb{RP}^2$. I claim that the degree of the map on the level of $S^2$ is a homotopy invariant of the map on the level of $\Bbb{RP}^2$. For pick a homotopy $f_t: \Bbb{RP}^2 \times I \to \Bbb{RP}^2$. By assumption $f_0$ came from an odd map $\tilde f_0: S^2 \to S^2$. I claim that there is a lift $\tilde f_t: S^2 \times I \to S^2$ such that all of the $\tilde f_t$ are odd.

This is pretty easy: just pick a lift! I claim that any lift is automatically odd. For if $\tilde f_t$ lifts $f_t$, we necessarily have $\tilde f_t(\{x,-x\}) = \{x,-x\}$; if this is sufficiently close to an odd map, then $\tilde f_t(x)$ must be close to $-x$; so $\tilde f_t(x)$ must actually be $-x$.

So any map that descends from an odd map $S^2 \to S^2$ is only homotopic to maps that descend from odd maps, and the homotopy class of the odd map is a homotopy invariant of the map $\Bbb{RP}^2 \to \Bbb{RP}^2$.

In particular, its degree is determined by the odd map up above. Now all you need to know is that there are odd maps of arbitrary odd degree. I rather believe I once proved this but I don't remember the construction right now. I'll edit it in if I remember it.

The same thing works for even maps - but for $S^2$ an even map must be degree 0. The lift of a map $\Bbb{RP}^2 \to \Bbb{RP}^2$ is either even or odd, so we've now classified all of them.

Too long for a comment:

@MikeMiller: you seem to suggest that the maps with even lift are homotopically trivial on $\mathbf P^2(\mathbf R)$, which is not the case, I think: let $\pi$ be the quotient map $S^2\rightarrow \mathbf P^2(\mathbf R)$ and let $g\colon\mathbf P^2(\mathbf R)\rightarrow S^2$ be the map that contracts the $1$-cell to a point. Let $\tilde f$ be the composition $g\circ \pi$. It is clearly an even map from $S^2$ to itself. Allthough it is of degree $0$, it is not homotopically trivial between all even maps from $S^2$ to itself. Indeed, otherwise $g$ would be homotopically trivial as well, which it is not; $g$ is the unique homotopically nontrivial map from $\mathbf P^2(\mathbf R)$ to $S^2$. It follows that the map $f$ from $\mathbf P^2(\mathbf R)$ to itself induced by $\tilde f$, i.e., $f=\pi\circ g$, is homotopically nontrivial.