= Solution
Let $a$ and $b$ be the pullbacks of the standard generator of $H^n(S^n;\mathbb Z)$ from the two factors. The <Künneth theorem> and graded commutativity of the <cup product> give
$$
\boxed{H^*(S^n\times S^n;\mathbb Z)
=\mathbb Z\langle1,a,b,ab\rangle,qquad
|a|=|b|=n,quad a^2=b^2=0,quad ba=(-1)^n ab.}
$$
Each homeomorphism acts invertibly on $H^n\cong\mathbb Z^2$. To respect ordinary composition, send $h$ to $(h^{-1})^*$; functoriality of <induced map on cohomology> then defines a homomorphism
$$
\operatorname{Homeo}(S^n\times S^n)\longrightarrow GL(2,\mathbb Z).
$$
For $n=1$, the space is the <torus>. Every matrix in $GL(2,\mathbb Z)$ induces a linear homeomorphism $\mathbb R^2/\mathbb Z^2\to\mathbb R^2/\mathbb Z^2$, so the image is all of $GL(2,\mathbb Z)$.
For $n=2$, write $h^*a=pa+qb$. Since $a^2=0$ and $ab=ba$,
$$
0=(h^*a)^2=2pq,ab,
$$
so $pq=0$; the same argument applies to $h^*b$. Invertibility then forces the matrix to be a <signed permutation matrix>. Every such matrix is realized by swapping the two sphere factors and applying an orientation-reversing homeomorphism to either factor. Thus the image consists exactly of the eight signed permutation matrices.
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