Show a function is well defined 
My understanding of a function being well defined:
Let $(G,\circ)$ be a group and let $H\unlhd G$ be a normal subgroup. Let $G/H=\{ah\mid a\in G\}$.
For a group $(G/H,\star)$(the quotient group of $G$ by $H$), where $aH\star bH:=(a\circ b)H$
To show the function $\star$ is well defined, we can show that if $aH=a'H$ and $bH=b'H$ for some $a',b'\in G$, then $(ab)H=(a'b')H$. This can be done from using the fact that $H$ is normal:
$(ab)H=a(bH)=a(b'H)=a(Hb')=(a'H)b'=a'(Hb')=(a'b')H$

Given: Let $G$ be a group acting transitively on a set $X$ (i.e. there is a single orbit),
let $S_G(X)$ be the set of functions $\sigma:X\rightarrow X$ that commute with the action of $G$ (i.e. $\sigma(g\cdot x)=g\cdot\sigma(x),\forall g\in G$ and $x\in X$)
let $x_0$ be an element of $X$ with stabilizer $G_{x_0}=H$
and for $n\in G$ define $\sigma_n:X\rightarrow X$ by $\sigma_n(g\cdot x_0)=(gn)\cdot x_0$ for $x\in X$
Question: Show that $\sigma_n$ is well-defined iff $n$ is an element of the normalizer $N_G(H)$ of $H$ in $G$
My answer: $N_G(H)=\{g\in G\mid g^{-1}Hg=H\}$, so the question can be reduced to: show $\sigma_n$ is well defined iff $n\in G$ is an element which commutes with every element of $H=\{g\in G\mid g\cdot x_0=x_0\}$
Point of contention: I have tried to write out a similar process as above for $\sigma_n$ being well defined, here is my attempt:
If $g\cdot x_0=g'\cdot x_0'$ and $h\cdot y_0=h'\cdot y_0'$ then $(g\cdot x_0)(h\cdot y_0)=(g'\cdot x_0')(h'\cdot y_0')$.
Replacing each element on the LHS, we get the RHS.
I feel that this must be wrong, how should the answer look, thanks in advance for the help.
 A: I'll start with a very useful way of considering well-definedness that a professor of mine once told me.
A map $$f:A\to B \atop \quad\quad a\mapsto f(a)$$


*

*Is well defined "on the right" if for each $a\in A$, the element (of some superset that I haven't specified here) $f(a)$ is indeed an element of $B$.

*Is well defined "on the left" if (informally spoken) the way we get $f(a)$ doesn't actually depend on how we need to "represent" $a$. In other words, it needs to be that $a = b \Rightarrow f(a) = f(b)$. In a lot of cases this is trivial, but as I'm sure you know, sometimes we need to make sure this is true (the typical case being that of a function defined on equivalence classes).


Here, "on the right" is obvious, so we need to look at the left side. The problem will be in our choice of $g$ in $(g\cdot x_0)$. The way the function is defined implicitly says that $\sigma_n(x) = (gn)\cdot x_0$, where $g$ is such that $x = g\cdot x_0$. You have to prove two things: 


*

*Such a $g$ always exists, for any $x\in X$

*If $x = g\cdot x_0 = g'\cdot x_0$, then $(gn)\cdot x_0 = (g'n)\cdot x_0$
