$$T^2=\{(e^{2\pi ix},e^{2\pi iy})\in \mathbb C^2|x,y\in \mathbb R\}.$$ Suppose $f,g:T^2 \rightarrow T^2:$ $$f\left(\left(e^{\displaystyle2\pi ix},e^{\displaystyle2\pi iy}\right)\right)=\left(e^{\displaystyle2\pi i(2x+3y)},e^{\displaystyle2\pi i(x+2y)}\right) $$
$$g\left(\left(e^{\displaystyle2\pi ix},e^{\displaystyle2\pi iy}\right)\right)=\left(e^{\displaystyle 2\pi i(2x+3y)},e^{\displaystyle 2\pi i(x+y)}\right) .$$
Prove $f$ and $g$ are both homeomorphism mapping of $T^2$ but $f$ is not homotopy equivalent with $g.$
You show that $f$ and $g$ act differently on $H_1(T)$. With respect to the usual basis, $f$ acts via $\pmatrix{2&3\\1&2}$ and $g$ acts via $\pmatrix{2&3\\1&1}$.