If you have the semi-ring structure, with $+$ etc., then every other nonstandard element is even and every other element is odd, so the types within a $Z$ chain are not identical.

If you just have the order, then the types are identical, since there are numerous automorphisms, translating within a $Z$-chain or moving various $Z$ chains.

Since the nonstandard numbers believe that every other number is even and every other number is odd, a fact that is expressible in the language you mention, it follows that the types are not the same for every two elements in a $Z$-chain. In fact, more is true: any two nonstandard natural numbers in a common $Z$-chain have distinct types, for if they differ by a finite number $n$, then they will have different residue modulo $n+1$, making their types different.

If one restricts attention only to the order, however, then any two elements in a nonstandard $Z$-chain have the same type, since there are order-automorphisms that shift within this $Z$-chain. And in a countable nonstandard model, the $Z$-chains are ordered like the rationals, and so the order automorphism group acts transitively on these elements. It follows that all the nonstandard elements have the same type in the language containing only the order.