The ground state of the spin-$1$ AKLT model gives an example of a $\mathbb{Z}_2 \times \mathbb{Z}_2$ symmetry-protected topological (SPT) phase, the Haldane phase. This state is a nice example of the Haldane phase because it has an exact matrix product states (MPS) representation. This state has exponentially decaying connected correlation functions.
Is there an example of a spin-$1$ state that's in the Haldane phase but has zero correlation length?
In particular, I'd want the protecting symmetry to be the same $\mathbb{Z}_2 \times \mathbb{Z}_2$ that protects the AKLT state (i.e. $\pi$ rotations about $x$, $y$, and $z$ axes), or the inversion symmetry that protects the AKLT state. I'm currently learning about SPT phases, so it's possible I'm missing or misunderstanding something trivial.
For example, in $1$d, there's the spin-$1/2$ cluster state, which is again an exact matrix product state protected by a $\mathbb{Z}_2 \times \mathbb{Z}_2$ symmetry. The cluster state is even nicer than the AKLT state, in a sense, in that its correlation functions vanish identically after a finite distance. That is, unlike AKLT, its correlation length is zero.
My motivation is that I'm curious about whether there exists something similar to the cluster state but for spin-$1$ particles, and still in the Haldane phase and protected by the same symmetries (or a subset of them) as the AKLT state. I found a paper that mentions a generalization of cluster states to qudits, but it looks like the symmetry group there is larger, $\mathbb{Z}_3 \times \mathbb{Z}_3$ for spin-$1$. Further, it's not clear to me that breaking that symmetry down to some $\mathbb{Z}_2 \times \mathbb{Z}_2$ would yield the same $\mathbb{Z}_2 \times \mathbb{Z}_2$ ($\pi$ rotations about $x$, $y$, and $z$) that protects the AKLT state.