= Solution
The <complex tautological line bundle> over <Complex projective space> is
$$
\lambda=\{(\ell,v):\ell\in\mathbb{CP}^n,\ v\in\ell\}\longrightarrow\mathbb{CP}^n,\qquad(\ell,v)\longmapsto\ell.
$$
On the standard chart $U_i=\{[z_0:\cdots:z_n]:z_i\ne0\}$, the vector $v_i([z])=z/z_i$ is independent of the chosen representative and gives a continuous nonzero frame. The map
$$
U_i\times\mathbb C\longrightarrow\lambda|_{U_i},\qquad(\ell,w)\longmapsto(\ell,wv_i(\ell))
$$
is a <local trivialization>; its inverse takes the $i$th coordinate of the vector in the fibre. Hence $\lambda$ is a locally trivial <complex line bundle>.
Its unit <sphere bundle> is $S^{2n+1}$: a unit vector $v\in\mathbb C^{n+1}$ corresponds to $([v],v)$. The projection is the <Hopf fibration>. Regard $\lambda$ as an oriented real rank-two <vector bundle> using its complex orientation, and put $t=e(\lambda)\in H^2(\mathbb{CP}^n;\mathbb Z)$. The <Gysin sequence of a sphere bundle> gives, for $2\leq q\leq2n$,
$$
H^{q-2}(\mathbb{CP}^n;\mathbb Z)\overset{\smile t}{\longrightarrow}H^q(\mathbb{CP}^n;\mathbb Z)
$$
as an isomorphism, since the intervening cohomology groups of $S^{2n+1}$ vanish. It also gives $H^1(\mathbb{CP}^n)=0$. The <CW complex> structure has one cell in dimensions $0,2,\ldots,2n$ and none above $2n$, so these groups and products give
$$
\boxed{H^*(\mathbb{CP}^n;\mathbb Z)\cong\mathbb Z[t]/(t^{n+1}),\qquad |t|=2.}
$$
The <Euler class> of $\lambda$ is the negative of the usual hyperplane generator; replacing $t$ by $-t$ gives the same ring presentation. For $n=0$ the formula reads $\mathbb Z$.
Now let $\alpha$ generate $H^2(S^2;\mathbb Z)$ and let $u_i$ be its pullback from the $i$th factor of $(S^2)^n$. The <Künneth theorem> gives $H^2((S^2)^n;\mathbb Z)=\bigoplus_i\mathbb Z u_i$. If a map $f$ were invariant under all factor permutations, write $f^*\alpha=\sum_i a_i u_i$; naturality under a transposition forces every $a_i$ to have the same integer value $a$. Let $\Delta:S^2\to(S^2)^n$ be the diagonal. Then
$$
\Delta^*f^*\alpha=na\alpha.
$$
The second condition is $f\circ\Delta=\mathrm{id}$, so this must equal $\alpha$, giving $na=1$. \b[For $n>1$ this is impossible in $\mathbb Z$, and $\boxed{\text{no such continuous map exists}.}$] This is the <diagonal-degree obstruction to a symmetric sphere retraction>.
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