= Solution
For a rank-$r$ real <vector bundle> with a <fiber metric>, the <Thom isomorphism theorem> says that if the bundle is oriented over the coefficient ring $R$, there is a <Thom class> $u\in H^r(D(E),S(E);R)$ restricting to the chosen generator on every fiber pair, and
$$
H^q(B;R)\xrightarrow{\ \pi^*(-)\smile u\ }H^{q+r}(D(E),S(E);R)
$$
is an isomorphism. For $R=\mathbb F_2$, every real vector bundle has the required orientation. The disk bundle retracts to $B$, so apply the pair long exact sequence and the Thom isomorphism to get the <Gysin sequence of a sphere bundle>
$$
\boxed{\cdots\to H^{q-r}(B;R)\xrightarrow{\smile e(E)}H^q(B;R)\xrightarrow{\pi^*}H^q(S(E);R)\xrightarrow{\pi_!}H^{q-r+1}(B;R)\xrightarrow{\smile e(E)}H^{q+1}(B;R)\to\cdots.}
$$
The first map is multiplication by the <Euler class of a vector bundle>, the pullback to the zero section of the absolute image of $u$. The connecting map followed by inverse Thom isomorphism gives $\pi_!$.
Apply this over $\mathbb F_2$ to the tautological real line bundle $\gamma$ on $\mathbb{RP}^n$. Its unit sphere bundle is $S^n$, with projection sending a unit vector to its line. Put $a=e_2(\gamma)=w_1(\gamma)\in H^1(\mathbb{RP}^n;\mathbb F_2)$, the <mod-two Euler class of a real line bundle>. For $n\geq1$ the degree-zero pullback $H^0(\mathbb{RP}^n)\to H^0(S^n)$ is an isomorphism. Exactness therefore makes the subsequent transfer zero, and multiplication by $a$ from $H^0$ into $H^1$ injective. In intermediate degrees the sphere cohomology vanishes, making each successive multiplication by $a$ an isomorphism. At degree $n$, the map from $H^{n-1}$ into $H^n$ is injective; the transfer $H^n(S^n)\to H^n(\mathbb{RP}^n)$ is surjective because $H^{n+1}(\mathbb{RP}^n)=0$. Since $H^n(S^n)=\mathbb F_2$, that top group is one-dimensional too, and $a^n$ is its nonzero generator. The same endpoint reasoning covers $n=1$. This gives the <Gysin proof of mod-two projective-space cohomology> and the <mod-two cohomology ring of real projective space>:
$$
\boxed{H^*(\mathbb{RP}^n;\mathbb F_2)=\mathbb F_2[a]/(a^{n+1}),\qquad|a|=1.}
$$
For $n=0$ this says simply $H^*(\mathbb{RP}^0)=\mathbb F_2$.
For a self-map of $\mathbb{RP}^4$, the induced map has $f^*a=\lambda a$ with $\lambda\in\mathbb F_2$, so $f^*a^j=\lambda^j a^j$. The mod-two alternating trace is
$$
L_2(f)=1+\lambda+\lambda^2+\lambda^3+\lambda^4=1
$$
for either value of $\lambda$. The alternating trace on homology equals that on the finite cellular chain complex; consequently this is the reduction modulo two of the integral <Lefschetz number>. That integer is nonzero, and the <Lefschetz fixed-point theorem> forces a fixed point. Thus \b[every self-map of the real projective four-space has a fixed point], a case of the <Lefschetz fixed-point property of even-dimensional real projective space>.
For the odd-dimensional example, let $J(x_1,x_2,x_3,x_4)=(-x_2,x_1,-x_4,x_3)$, so $J^2=-I$. Its projectivization is
$$
\boxed{F([x_1:x_2:x_3:x_4])=[-x_2:x_1:-x_4:x_3].}
$$
It is well-defined and continuous on $\mathbb{RP}^3$. A fixed projective point would have $Jv=tv$ for a nonzero real vector and real $t$, forcing $-v=J^2v=t^2v$ and therefore $t^2=-1$, impossible. This is a <fixed-point-free complex-structure map on odd-dimensional real projective space>.
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