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Let $x,y\in H^n(S^n\times S^n;\mathbb{Z})$ be the pullbacks of the two orientation classes. The cohomology ring of a product of two spheres hasand $xy$ generates $H^{2n}\cong \mathbb{Z}$. WriteThe matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ lies in $G{L}_2(\mathbb{Z})$ because a homeomorphism induces a cohomology ring automorphism.

If $n$ is even, graded commutativity givesThus $ab=cd=0$. Invertibility forces $A$ to be diagonal or anti-diagonal, and its two nonzero entries must each be $\pm 1$. Hence there are exactly eight possible actions: the signed permutation matrices.

Evenness is necessary. For $n=1$, the product is the torus, and every $A\in G{L}_2(\mathbb{Z})$ is induced by an integral linear automorphism of the torus. For example, the infinitely many matrices $\left(\begin{array}{cc}1& m\\ 0& 1\end{array}\right)$ give distinct actions on $H^1$.

Let $A\in G{L}_2(\mathbb{Z})$. If $r-r^{\prime }\in {\mathbb{Z}}^2$, then $A(r-r^{\prime })\in {\mathbb{Z}}^2$, sodefines a continuous map on $X={\mathbb{R}}^2/{\mathbb{Z}}^2$. Since $\det A=\pm 1$, the integral general linear group contains $A^{-1}$ as another integer matrix, and $f_{A^{-1}}$ is the inverse of $f_A$. Hence $f_A$ is a homeomorphism, and $f_A(x_0)=x_0$.

The map $A:{\mathbb{R}}^2\to {\mathbb{R}}^2$ is the lift of $f_A$ fixing the origin. A loop whose lift ends at $m\in {\mathbb{Z}}^2$ is sent to a loop whose lift ends at $Am$. Under the isomorphism in part (a), the induced map is thereforeThis is precisely the integral linear automorphism of the torus associated with $A$.