Let $G$ be a group acting on a set $X$, and let $H$ be a group acting on a set $Y$. I'm wondering if there is standard terminology or notation to express the following condition: there exist an isomorphism $\varphi:G\rightarrow H$ and a bijection $\sigma:X\rightarrow Y$ such that $$\sigma(gx)=\varphi(g)\sigma(x)$$
for all $g\in G$ and all $x\in X$. Basically, I want to say that not only are $G$ and $H$ isomorphic as abstract groups, but that they act in the same way on the corresponding sets.
[Math] Isomorphism of group actions
group-theory
Related Solutions
Let us remember the definition of faithful, transitive and free:
The action of a group $G$ on a set $X$ is called:
Transitive: if $X$ is non-empty and if for any $x,y\in X$ there exists $g\in G$ such that $g\cdot x = y$.
Faithful (or effective): if for any two distinct $g,h\in G$ there exists $x\in X$ such that $g\cdot x \neq h\cdot x$; or equivalently, if for any $g\neq e\in G$ there exists $x\in X$ such that $g\cdot x \neq x$. Intuitively, in a faithful group action, different elements of $G$ induce different permutations of $X$.
Free: if, given $g,h\in G$, the existence of $x\in X$ with $g\cdot x = h\cdot x$ implies $g = h$. Equivalently: if $g$ is a group element and there exists an $x\in X$ with $g\cdot x = x$ (that is, if $g$ has at least one fixed point), then $g$ is the identity.
Now we can determine your three group actions:
The action of $S_n$ on a set of $n$-elements is as you showed transitive. It is also faithful as you explained, but it is not free, because there are elements on $S_n$ that do have fixed points that are not the identity, for example $1\to 1, 2\to 3, 3\to 4, n-1\to n, n\to 2$. Since the action is transitive the group of orbits is the group of one element.
The action of $O(n)$ on $S^{n-1}$, it is transitive, since there is always a rotation that can send any point of $S^{n-1}$ onto the north pole. For $n>2$, It is not free since for any point on $S^{n-1}$ there are many rotations different from the identity, whose axis of rotation is the line that passes through that point and the origin, and all of these fix that point. It is not faithful neither since any element of $O(n)$ that is not the identity moves $S^{n-1}$, therefore there exists a point on $S^{n-1}$ that is not fixed by the action. Since the action is transitive the group of orbits is the group of one element. The case $n=2$ the action it is transitive, free but not faithful.
The action of $O(n)$ on $\mathbb{R}^n$ is not free since every element of $O(n)$ fixes the origin but is not necessarily the identity. It is not transitive either, since for two elements of $\mathbb{R}^n$ whose length are not equal, you cannot find an element of $O(n)$ that sends one on the other (the elements of $O(n)$ preserve norm). It is not faithful neither since when you restrict the action to $S^{n-1}$ it is not faithful either. Since $\mathbb{R}^n=[0,\infty)\times S^{n-1}$ the set of orbits can be identified with $[0,\infty)$ since when $O(n)$ is restricted to the second factor it is transitive.
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Yes. There might be multiple bijections. As a simple example, look at
$$\mathbb{Z} / 2 \times \mathbb{Z} / 2 \cong \mathbb{Z} / 2 \times \mathbb{Z} / 2$$
We can use the identity, which is an isomorphism. We can also use the flippy isomorphism $(a,b) \mapsto (b,a)$ (which is its own inverse).
As for your second question, I'm not sure what you mean by whether the "kind of isomorphism picked does not affect how the structural similarity between $A$ and $B$ [is] preserved"...
I will say that no matter what isomorphism you use, all the group-theoretic properties will be the same. For instance, their orders, how many subgroups, whether there exist torsion elements, etc. So I'll say that as a good first answer: no, it doesn't matter at all which isomorphism you use.
Of course, like everything in math, this is only mostly true. In fact, different isomorphisms carry different information, and just because two groups are isomorphic doesn't always mean we want to consider them the same! A good example of this is the study of lattices in $\mathbb{R}^n$. These are all (abstractly) isomorphic to $\mathbb{Z}^n$, but it turns out to be a bad idea to identify them all as "the same".
There are also times when different isomorphisms carry different information. A famous example is the isomorphism of a (finite dimensional) vector space to its double dual. There are lots of ways to do this (abstractly) but it turns out there's one unique best choice of isomorphism. To formulate this properly we need the language of category theory, and I'm not sure if this is the right place to go into that. Rest assured, though, you'll hear more about it as you continue progressing in mathematics!
The long and the short of it is that every isomorphism will preserve all the group theoretic properties of your object of interest. However different choices of isomorphism can sometimes lead to counter intuitive statements (galois theory is rather infamous for this...). If there is a canonical way to make the isomorphism, rather than finding some random isomorphism by making arbitrary choices along the way (choosing a basis, etc.) that canonical isomorphism tends to be "better behaved" in a way that you'll learn about when you start learning about categories.
I hope this helps ^_^
Best Answer
Let $X$ and $Y$ be a left $G$-set and a left $H$-set, respectively. Then, a homomorphism of group actions is a pair $(\varphi,\sigma)$, with $\varphi:G\to H$ being a group homomorphism and $\sigma:X\to Y$ being a function compatible with the group actions in the sense that $$\sigma(g\cdot x)=\varphi(g)\cdot\sigma(x)$$ for all $g\in G$ and $x\in X$. See for example Definition 2.1 on Page 115 of Groups and Computation II by L. Finkelstein and W. M. Kantor. What you have is just an isomorphism of group actions.
If $G=H$ and $\varphi=\text{id}_G$, then $\sigma:X\to Y$ satisfying $$\sigma(g\cdot x)=g\cdot \sigma(x)$$ for all $g\in G$ and $x\in X$ is usually called a homomorphism of (left) $G$-sets. In other words, if $\sigma:X\to Y$ is a $G$-set homomorphism, then $\left(\text{id}_G,\sigma\right)$ is an example of a homomorphism of group actions from $(G,X)$ to $(G,Y)$.