 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

Each homeomorphism acts invertibly on $H^n\cong {\mathbb{Z}}^2$. To respect ordinary composition, send $h$ to $(h^{-1}{)}^{\ast }$; functoriality of induced map on cohomology then defines a homomorphism

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 $h^{\ast }a=pa+qb$. Since $a^2=0$ and $ab=ba$,so $pq=0$; the same argument applies to $h^{\ast }b$. 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.