Consider a map $f: RP^n \rightarrow RP^n$ for n even. I want to show that $f$ has a fixed point by using the Lefschetz fixed point theorem. So we have to show for the Lefschetz number $l(f) \neq 0$. This follows from the fact that the induced map of $f$ on the zero-homology has to be the identity (thus the trace of its matrix has to be $1$), but I do not see why this has to hold.
Since $n$ is even $H_0RP^n=Q$ and for $k\neq0$ $H_kRP^n=0$, hence it is clear that all other induced maps are zero.
For any non-empty path-connected space $X$, $H_0(X)\cong\Bbb Z$, and any continuous map $f:X\to X$ induces the identity on $H_0(X)$.
The generator of $H_0(X)$ is the class of the zero-cycle $[P]$ for any point $P\in X$ (the cycle $[P]-[Q]$ is always a boundary in $X$). Thus $f_*$ takes $[P]$ to $[f(P)]$, which is homologous to $[P]$.