Unless I misundersand the notation very badly, this is surely not true. $\diamond$ is a finite lattice, and so the fixed point lattice of any $f : \diamond^n \to \diamond^n$ must also be finite, and so it must have a maximal element (the join of all the finite elements). Now, the infinite vertical lattice $(\mathbb{N}, \leq, \sqcap = \min, \sqcup = \max)$ has no maximal element, and so it can't be isomorphic to the fixed point lattice of some $f : \diamond^n \to \diamond^n$.
EDIT: I think I did misunderstand the notation -- maybe $i$ and $j$ aren't constants, but variables ranging over some ordered set $I$. So a cardinality argument still works, by taking the powerset of $\Sigma n:\mathbb{N}.\;\diamond^n \to \diamond^n$, which yields a lattice which has too many elements to be in isomorphism with the fixed point of any $f$.
Unless I misundersand the notation very badly, this is surely not true. $\diamond$ is a finite lattice, and so the fixed point lattice of any $f : \diamond^n \to \diamond^n$ must also be finite, and so it must have a maximal element (the join of all the finite elements). Now, the infinite vertical lattice $(\mathbb{N}, \leq, \sqcap = \min, \sqcup = \max)$ has no maximal element, and so it can't be isomorphic to the fixed point lattice of some $f : \diamond^n \to \diamond^n$.