For an integer $r\geq 1$, one can equip ${\mathbb N}^r$ with the product lattice structure defined by $(x_1,x_2,\ldots,x_r) \leq (y_1,y_2,\ldots,y_r)$ iff $x_k\leq y_k$ for every $k$. Similarly for ${\mathbb N}^{(\infty)}$, the set of all nonnegative integer sequences that are eventually zero (which can be viewed as the union of all the ${\mathbb N}^r$).
If $\sigma$ is any permutation of ${\mathbb N}^*$, the map $(x_1,x_2,\ldots) \mapsto (x_{\sigma(1)},x_{\sigma(2)},\ldots)$ defines a lattice automorphism (i.e. an order-preserving bijection) of ${\mathbb N}^{(\infty)}$.
Are there other automorphisms of ${\mathbb N}^{(\infty)}$ ?
No. Every automorphism of a lattice permutes its atoms, which here are the elements $(e_i)_j = \delta_{ij}$. And an automorphism of $\mathbb{N}^{(\infty)}$ is completely determined by how it permutes atoms, because every element can be defined order-theoretically starting from the atoms in terms of covering. I don't know a clean way to describe the full induction but e.g. $(2, 0, 0, \dots)$ is the unique element which covers $e_1$ but does not cover $e_2, e_3, \dots$, and $(2, 1, 0, \dots)$ is the unique element which covers the previous element but doesn't cover $(0, 2, 0, \dots), e_3, \dots$ and so forth. Generally we can induct on $\sum x_i$ to show that each element in a given degree is completely determined by how it covers elements of the previous degree.