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Jan 30, 2022 at 6:16 comment added Tim Campion Note that if you allow your structure to have two sorts, you can modify the example to have $\{a,b\}$ be the carrier for one sort and $\{1,2,\dots\}$ be the carrier for a second sort. Then I think you get that $G(\mathfrak A)$ in Noah's sense is precisely $\mathcal K$ (where all of the $f_G$'s go from the second sort to the first)
Dec 30, 2021 at 21:18 history edited Keith Kearnes CC BY-SA 4.0
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Dec 30, 2021 at 9:38 history edited Keith Kearnes CC BY-SA 4.0
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Dec 30, 2021 at 6:53 comment added Peter LeFanu Lumsdaine Just to clarify: this answer’s argument “the variable-appearance condition doesn’t restrict” shows that “for any algebra $A$, there’s an expanded algebra $A'$ such that the symmetry-groups of Noah-restricted terms of $A'$ are the symmetry groups of arbitrary terms of $A$” . But it doesn’t imply the converse, as Noah’s first comment here shows.
Dec 27, 2021 at 4:15 comment added Noah Schweber I've added a more focused subquestion to hopefully clarify the sort of thing I'm looking for.
Dec 26, 2021 at 23:58 history edited Keith Kearnes CC BY-SA 4.0
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Dec 26, 2021 at 23:51 history edited Keith Kearnes CC BY-SA 4.0
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Dec 26, 2021 at 23:28 comment added Noah Schweber This is interesting but I don't think it answers my question. Shifting to terms which depend on all variables changes the situation quite a bit: for the $\mathbb{G}(\mathfrak{A})$s as defined in my question, having a single nontrivial group automatically forces us to have groups of arbitrarily large finite order. And dropping the variable-appearance restriction **does** change the question: if we don't require every variable to actually appear in each term, then $\mathbb{G}(\mathfrak{A})$ must be closed under taking direct products with $S_n$s via dummy variables.
Dec 26, 2021 at 23:10 history edited Keith Kearnes CC BY-SA 4.0
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Dec 26, 2021 at 22:54 history answered Keith Kearnes CC BY-SA 4.0