Let $A$ be a finite set and let $\sigma:A^\mathbb{Z}\to A^\mathbb{Z}$ be the shift homeomorphism. Let $T:A^\mathbb{Z}\to A^\mathbb{Z}$ be a bijective homeomorphism that commutes with $\sigma$ (i.e. for every $x\in A^\mathbb{Z}$, $T\circ\sigma(x)=\sigma\circ T(x)$). By the Curtis-Hedlund-Lyndon theorem, $T$ is defined by some mapping $t:A^I\to A$, where $I\subseteq \mathbb{Z}$ is a finite interval, so that for any $x\in A^\mathbb{Z}$ and $n\in \mathbb{Z}$ we have $$T(x)_n=t(\sigma^{-n}(x)\upharpoonright I).$$
Choose $t$ so that the interval $I$ has the minimal possible length and denote by $|T|$ this length. Now since $T$ is invertible, $T^{-1}$ is again a homeomorphism commuting with $\sigma$, so we can define $|T^{-1}|$ in the same way.
My question is, how does $|T^{-1}|$ depend on $|T|$? Can it be arbitrarily larger, or there is some bound?
Yes, $|T^{-1}|$ can be arbitrarily larger than $|T|$ when $|A|$, the size of the alphabet, grows. However, in dimension 1 (the case considered in the question) there is a polynomial upper bound, see: upper bound and examples and quadratic upper bound for |T|=2.