Can a zero entropy automorphism of the torus have a Li-Yorke pair? If a continuous map of a compact metric space has positive topological entropy, then it has a Li-Yorke pair. In general, the converse is not true, and a zero entropy continuous map can have a Li-Yorke pair.
Can a zero entropy automorphism of the torus have a Li-Yorke pair? More generally, can a zero entropy endomorphism of a compact Lie group have a Li-Yorke pair? 
 A: I'm assuming a Li-Yorke pair for $T$ is a pair $x,y$ such that $\limsup_{n\to \infty} d(T^n(x), T^n(y)) > 0 $ and $\liminf_{n\to \infty} d(T^n(x), T^n(y)) = 0$.
For toral automorphisms, I think the answer to your question is no. The idea is this: if $T$ is ergodic, then it has positive entropy (every ergodic linear toral automorphism does, see for instance here). If $T$ is not ergodic, you can find a factor of (some power of) $T$ on a torus of lower dimension which is ergodic, which means again that $T$ has positive entropy.
Indeed, being non-ergodic implies that there is an eigenvalue which is a root of the unity. We may replace $T$ by some power of $T$ and assume that $1$ is an eigenvalue. Moreover you can find an integer eigenvector $v$ with eigenvalue $1$ (because the coefficients of the matrix corresponding to $T$ are integer), so passing to the quotient by $\mathbb{R}v$ you should get a torus of dimension $n-1$. Since $T$ had a Li-Yorke pair, so does the automorphism induced by $T$ on the quotient, so you have reduced the dimension. If the new map is still not ergodic, we may repeat this reduction. But this process has to stop before reaching dimension $1$ since you cannot have a Li-Yorke pair in dimension $1$.
Note: I edited this answer since the previous one had a big mistake (the claim was false)
