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I do hope it's appropriate to post all the random questions I have here... Specifically I'm thinking about applying group theory to some "universal set".

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Sure! But technically such a thing should be called something different to distinguish it from the set case; I'd personally use "group class-action" for a set-sized group acting on a class, and "class group action" for a class-sized group acting on a class.

One simple (if not very interesting) example of a group class-action is the group $\mathbb{Z}/2\mathbb{Z}$ acting on $ON$, the class of ordinals, by having the nontrivial element permute each "odd" ordinal $2\alpha+1$ with its predecessor $2\alpha$ (so e.g. $\omega$ would be sent to $\omega+1$, and vice versa, by the action of the nontrivial element).

An interesting example of a class group-action would be the class group of all permutations of $ON$ with set-sized support acting on $ON$ in the natural way. (We need set-sized support to view this group as a class - otherwise, each element of the group is a class function, and the group itself is some kind of "hyperclass"!)

And in fact they're useful in abstract algebra: class actions can represent very complex groups! Sorry, I couldn't resist.

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    For example: https://www.iclg.co.uk/practice-areas/class-and-group-actions/class-and-group-actions-2017 – 2017-02-23
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    @Noah So could, say **G**(ξ,★) (where ξ is the Class of all sets (or perhaps even all mathematical objects)), exist? – 2017-02-23
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    @Sup I don't know what $G(\xi, \star)$ means, but: probably! Incidentally, the class of all sets is more commonly denoted by ["$V$"](https://en.wikipedia.org/wiki/Von_Neumann_universe). – 2017-02-23
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    @Noah is that Conway's On/No you're referring to? I really need to get back into ONAG :p and I'm just using G to denote the group, and ★ some arbitrary operation on ξ (or I guess V) – 2017-02-23
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    @Sup No, I'm using "$ON$" to denote the class of ordinals: "acting on $ON$, the class of ordinals." Sometimes "$Ord$" is used instead. And yes, the object $G(V, \star)$ exists, but it is a *class* (namely, the class of ordered triples $(g, x, y)$ with $g\in G$ and $x, y$ sets, where each such triple is understood as meaning "$g(x)=y$"). – 2017-02-23
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    @Noah "No" - I see what you did there. Thankyou for your help! Is it feasible to find ★..? – 2017-02-23
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    @Sup I don't know what you mean - there are many possible class actions of a group on a class. So there's no unique $\star$. (Oh, small abuse of notation in my previous comment: by "$g(x)=y$" I meant "$g\star x=y$".) – 2017-02-23
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    E.g. is there an operation that can act meaningfully on any mathematical object? Something I can do to a number, or a set, or a shape, or a whatever... I may just be too ill-experienced to tackle such questions at this stage :L – 2017-02-23
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    @Sup Yes, *like I said, there are many such actions*. For instance, the trivial action, or the action of $\mathbb{Z}/2\mathbb{Z}$ where the nontrivial element swaps $\{\}$ and $\{\{\}\}$ and leaves all other sets fixed, or many others. There is not a unique action of any (nontrivial) group on $V$. (cont'd) – 2017-02-23
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    Note, however, that I don't think you've asked quite what you mean: I think what you're asking for is a group which acts on each set (possibly of a certain type), rather than a group which acts *on the class of sets itself). E.g. I think you want a group which acts on every topological space, not a group that permutes topological spaces. In this case, you have to do a lot of work to formalize what you mean, but essentially however you do it, your action will be the trivial action on "most" spaces. – 2017-02-23
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    +1 + (+1) for the final joke. – 2017-02-23
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    @AndreasCaranti I heard a story about a number theorist who had business cards made that said "Representing groups in all fields". – 2017-02-23