= Solution
Use cohomological degree, and write the <Sullivan minimal models> as
$$
M(\mathbb{CP}^n)=(\Lambda(x_2,y_{2n+1}),d),
\qquad dx=0,\quad dy=x^{n+1},
$$
and
$$
M(S^{2n})=(\Lambda(u_{2n},v_{4n-1}),d),
\qquad du=0,\quad dv=u^2.
$$
The subscripts indicate degrees. The even generators generate a <polynomial ring> and the odd generators an <exterior algebra>. Their <cohomology rings> are respectively $\mathbb Q[x]/(x^{n+1})$ and $\mathbb Q[u]/(u^2)$, since each odd differential kills exactly the indicated non-zero-divisor polynomial relation. These are the usual projective-space and even-sphere <minimal models>.
Orient the top classes so that the degree-one condition is $f^*[u]=[x^n]$. A representative model map must send $u$ to $x^n$. It must send $v$ to an element of degree $4n-1$ with differential $x^{2n}$. The element $x^{n-1}y$ has exactly these properties. Thus
$$
\boxed{f^*:M(S^{2n})\longrightarrow M(\mathbb{CP}^n),
\qquad u\longmapsto x^n,\quad v\longmapsto x^{n-1}y.}
$$
The differential check is $d(x^{n-1}y)=x^{n-1}x^{n+1}=x^{2n}=(f^*u)^2$. In these degrees there is no other available monomial: even powers of $x$ have even degree, and an odd monomial is a power of $x$ times $y$. Hence the coefficient is forced by the differential. For $n=1$ the two models coincide and the map is the identity model map.
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