= Solution
Work with coefficients $\mathbb F_2$. For $n\ge1$, the <mod-two cohomology ring of real projective space> and the <Künneth theorem> give
$$
H^*(\mathbb{RP}^{2n};\mathbb F_2)=\mathbb F_2[z]/(z^{2n+1}),
\qquad
H^*(\mathbb{RP}^{n}\times\mathbb{RP}^{n};\mathbb F_2)
=\mathbb F_2[x,y]/(x^{n+1},y^{n+1}),
$$
where all three generators have degree one. Any <induced map on cohomology> has $f^*z=\alpha x+\beta y$, with $\alpha,\beta\in\mathbb F_2$. By naturality of the <cup product>, the only possible term in top degree is
$$
f^*(z^{2n})=(\alpha x+\beta y)^{2n}
=\binom{2n}{n}\alpha^n\beta^n x^ny^n.
$$
All other terms vanish because their exponent of $x$ or $y$ exceeds $n$. But
$$
\binom{2n}{n}=2\binom{2n-1}{n-1}
$$
is even for every $n\ge1$. Thus the displayed top-degree <induced map on cohomology> is zero for every $f$.
Both top-degree <homology groups> are one-dimensional over $\mathbb F_2$, and their dual <cohomology groups> pair with them nondegenerately. Naturality of this evaluation makes $f_*$ dual to $f^*$, so $f_*$ is also zero and cannot be an <isomorphism>. This is the <mod-two degree obstruction for equal projective factors>. Consequently
$$
\boxed{\text{There is no such map for any positive }n.}
$$
If zero is included among the allowed values, $\mathbb{RP}^0$ is a point and the unique map is an <isomorphism> on $H_0$. In that convention, \b[the only value is $n=0$].
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