I've always liked lattice QFT because it's mathematically unambiguous and non-perturbative, but it does have two drawbacks: (1) the lattice is artificial, and (2) some features are messy. One of those messy features is chiral symmetry. As an example, the lattice model of a free Dirac fermion in even-dimensional spacetime is easy to construct, but if we construct it in such a way that the non-chiral symmetry $\psi\to\exp(i\theta)\psi$ is on-site, as we usually do, then the Nielsen-Ninomiya theorem tells us that the chiral symmetry $\psi\to\exp(i\theta\Gamma)\psi$ cannot be on-site. (I'm using $\Gamma$ for the product of all the Dirac matrices.)
Instead of thinking of that messiness as a drawback, we can also think of it in a more positive way: it almost helps us understand why the chiral anomaly must exist. On-site symmetries can always be gauged, but we have no such guarantee for non-on-site symmetries. If that last part could be strengthened to "non-on-site symmetries cannot be gauged," then this really would help us understand why the chiral anomaly must exist.
Question: It is true that non-on-site symmetries cannot be gauged? In other words, is a gaugeable symmetry necessarily on-site in every lattice version of the theory?
I'm guessing that this is not true. I don't know why we'd ever want to use a lattice version in which a gaugeable symmetry group is realized as a non-on-site symmetry, because that would be unnecessarily messy,$^\dagger$ but is it actually impossible?
My question is similar to another unanswered question, but that one asks about ensuring the absence of anomalies, while I'm asking about ensuring the presence of an anomaly.
$^\dagger$ I say "unnecessarily" because I'm assuming that anomaly-free symmetries can always be on-site in some lattice version. I don't know if that's true, either. That probably-more-difficult question has already been asked on Physics SE.
Also related: When can a global symmetry be gauged?