What is a "natural group action"?

Question

Eg. The symmetric group on S acts on S in a natural way, for all sets S.

Thanks in advance!

Answer 1

It's the action $Sym(S) \times S \to S$ given by $(f,x) \mapsto f(x)$.

Every action $G \times S \to S$ defines a homomorphism $G \to Sym(S)$ and vice-versa. The natural action defined above corresponds the identity homomorphism. This makes it "natural".

An "unnatural" group action $Sym(S) \times S \to S$ could for instance correspond to an nontrivial automorphism of $Sym(S)$ induced by renaming the elements of $S$.

Answer 2

For comparison's sake, we can try to build an "unnatural" group action. Given a group $G$ and a set $S$, we know that the identity $e \in G$ must act as the identity map and that the action of two elements is the same as the action of their composition.

I propose as our "unnatural action" we take $S_n$ as our group and $S$ to be a set of $n$ labeled elements. Then, instead of $(1,2) \in S_n$ transposing the $1^{st}$ and $2^{nd}$ elements, it transposes the $5^{th}$ and $9^{th}$ elements. There is certainly a way to make this a well-defined group action, but it doesn't feel natural at all.