Here's a counterexample: on Z^2, f(x,y)=(x,y+x). More generally, the order-preserving automorphisms of Z^n are exactly the upper triangular matrices with 1s on the diagonal (this should be easy to see by combining Charles's argument with my example in the case n=2, and then the generalization to arbitrary n isn't too hard).

Here's a counterexample: on $\mathbb{Z}^2$, $f(x,y)=(x,y+x)$.

More generally, the order-preserving automorphisms of $\mathbb{Z}^n$ are exactly the upper triangular matrices with 1s on the diagonal (this should be easy to see by combining Charles's argument with my example in the case $n=2$, and then the generalization to arbitrary $n$ isn't too hard).

Here's a counterexample: on Z^2, f(x,y)=(x,y+x).

Here's a counterexample: on Z^2, f(x,y)=(x,y+x). More generally, the order-preserving automorphisms of Z^n are exactly the upper triangular matrices with 1s on the diagonal (this should be easy to see by combining Charles's argument with my example in the case n=2, and then the generalization to arbitrary n isn't too hard).