Is a constant transformation still considered a gauge transformation? I've never even considered the possibility that a constant transformation would not qualify as a gauge transformation.  But I'm reading a paper that seems to make exactly this distinction.  In particular, the title of the paper itself begins with "Gauge Invariant".  But their results clearly change under any Poincaré transformation (or more generally any BMS transformation).  They even acknowledge this at one point deep in the paper.  The context is gravitational radiation on $\mathscr{I}^+$.
Now, they have eliminated a more dynamical form of gauge invariance, which can vary from point to point.  But I would say that at least somewhere early on in the paper, they should clarify that the "Gauge Invariant" they use in the title and throughout the paper only refers to those parts of the gauge freedom.  In my opinion, an unqualified "gauge invariant" necessarily refers to all possible gauge transformations.
Am I being over-precise?  Would most people normally understand that "gauge invariant" excludes constant gauge transformations?
 A: There are such things are large gauge transformations, which I think are related to your question. For example, consider general relativity where the gauge invariance is diffeomorphism invariance. Typically gauge transformations are considered that leave the boundary invariant, but there are also large gauge transformations that for example rescale the boundary metric. As an example, $t \rightarrow 2t$ is technically a gauge transformation but it changes the asymptotic value of the metric. Similar statements hold in gauge theories as well.
Large gauge transformations can affect conserved charges, so although these are technically gauge transformations, they are often distinguished from "small" or what we might call ordinary gauge transformations.  
A: In YM theory, you generally don't take the constant transformations to be gauge transformations, since the constant transformations are generated by the charge operator.  If the charge operator generated gauge transformations, it would act trivially on all physical states, which would mean that you couldn't have charges.
A: I can't think of any sane definition of the phrase "gauge transformation" which would exclude a function that is identically equal to some constant $C$, but include a function that is equal to $C$ everywhere except for at a tiny region over in the Andromeda galaxy, where it takes on the value $C + \epsilon$ for a tiny shift $\epsilon$. So yes, a constant transformation still counts as a gauge transformation.
