This is a homework question for an introduction to group theory class.
Here is the preamble of the homework: Let $G$ be a group and let $X$ and $Y$ be two sets. Suppose that $G$ acts on $X$ and $Y$. A function $f:X\rightarrow Y$ is called G-equivariant provided that for all $g ∈ G$ and $x ∈ X$, one has $f(g ·x) = g ·f(x)$. A G-equivariant isomorphism between $X$ and $Y$ is a G-equivariant function $f : X → Y$ such that there exists a G-equivariant function $g : Y → X$ where $f◦g=1_Y $ and $g◦f=1_X$.
Here is the question restated: Let $G$ act on $X$, and let $x ∈ X$ be given. Prove that there is a unique action of $G$ on ${Orb}_G (x)$ such that the inclusion ${Orb}_G (x) \hookrightarrow X$ is $G$-equivariant.
Here is my work: We are asked to find a unique action of $G$ on ${Orb}_G (x)$. We are already given that $G$ acts on $X$. Let's quickly check that this means that our inclusion map is $G$-equivariant.
$\iota (g\cdot x)=g\cdot x=g\cdot \iota (x)$ and so the inclusion map is $G$-equivariant here.
Suppose $G$ acts on ${Orb}_G (x)$ in another way but keeping the necessary properties from the question asked. Let's denote this action with $\star$. Then it suffices to show that $g\star x =g\cdot x$.
$g\star x=\iota (g\star x)=g\star \iota(x)$ ??
Well we know $\iota$ is $G$-equivariant with respect to either action. So
$\iota ((g\cdot h)\cdot x)=(g\cdot h)\cdot \iota(x)$
$\iota ((g\star h)\star x)=(g\cdot h)\star \iota(x)$
Ok another dead end.
Maybe we use the Orbit-Stabilizer theorem in some way? I don't think it would make sense to use it in this context. Perhaps considering a permutation representation would make more sense to try?
$\rho (g)(x)=g\cdot x$ is the notation used to denote the so-called permutation representation. But switching the notation here doesn't seem to make it clear where to keep going?
The previous question asked us to show that a $G$-equivariant function $f:X\rightarrow Y$ is bijective if and only if it is an isomorphism which I have already proven on my own. But I'm not sure if this is particularly illuminating. Perhaps we could consider the restriction of the codomain so the inclusion is then just the identity and work out something through there?
Anyways, I am a bit stumped and would appreciate any insights on how to proceed.
You had it, but you got confused. The best way to not get confused is to get back to the definition of group actions: It is a morphism $\rho : G \to \mathfrak{S} (X)$. (I denote by $[x]$ the orbit of $x$)
Let us show that the only possible action is the one given by restriction, or by precomposing the inclusion: We set $\imath: [x] \to X$ the inclusion. The restiction action $\rho'$ is given by $\rho' (g) = \rho(g)_{|[x]}$. Which is well defined since for any $y \in [x]$, $\rho(g)(y) \in [y] = [x] $.
First of all, it is quite obvious that it works, take $y \in [x], \, g \in G$: $$\rho'(g) \circ \imath(y) = \rho'(g)(y) = \rho(g)_{|[x]} (y) = \rho (y) = \imath \circ \rho(g) (y)$$ so it is equivariant.
Now suppose that you have another group action $\rho'': G \to \mathfrak{S}([x])$, if $\imath$ is equivariant, $$\rho''(g) (y) = \imath \circ \rho''(g)(y) = \rho(g) \circ \imath (y) = \rho(g) (y) = \rho'(g) (y)$$ so it is unique.
Edit Why do we have $\imath \circ \rho' = \rho'$ (and equivalently for $\rho''$)?
It depends on the pot you want to have, either your work with groups and group actions by looking at what it means for its elements. If you have an inclusion $\imath A \to B$ where $A \subseteq B$, $\imath(a) = a$ for $a \in A$. So here since $\rho'(g) (y) \in [x]$, you get $\imath \circ \rho'(g) (y) = \rho'(g) (y)$!
Why do we bother with inclusion at all? It is important because we want to be rigorous with the sets/objects we are working in: $a \in A$ and $\imath(a) \in A \subseteq B$ and there is a moral difference: in the first case we look at $a$ as an element of $A$ and in the second case, we look at it as an element of $B$ which fortunately happens to also be an element of $A$, but we see it as an element of $B$ verifying a certain property instead of an element of $A$. It also allows to formally use composition of maps in order for it to be well defined.
Here is a more "categorical friendly proof". Inclusions are monomorphism, this means that for any adequate maps $f,g$, $\imath \circ f = \imath \circ g \implies f = g$. Here this gives us (only with equivariance)
$$\imath \circ \rho'(g) = \rho(g) \circ \imath = \imath \circ \rho''(g)$$
thus $\rho'(g) = \rho''(g)$!