Some Preliminary Definitions and Properties:
Actions of a group $G$ on sets $X$ and $Y$ are equivalent if the corresponding action of $G$ on maps from $X$ to $Y$ fixes some bijection. In this case, we write $X \sim Y$.
An action is said to be transitive if it has only one orbit.
Given an action of a group $G$ on a set $X$,
the stabilizer, $G_x$, of $x$ is defined by
$$ G_x = \{ g \in G : gx = x \}; $$
and the orbit, $Gx$, of $x$ is defined by
$$ Gx = \{ gx : g \in G \}. $$
Properties:
(i) $Gx \subset X$ and $G_x \le G$;
(ii) there exists a bijection from $Gx$ to the coset space $G/G_x$; and
(iii) if $G$ is finite, then $|Gx|=|G/G_x|$.
The Statement of the Problem:
Prove the following:
(a) For each subgroup $H$ of a group $G$, the action $G$ by left translation on the left coset space $G/H$ is transitive.
(b) Each transitive action of a group $G$ on a set $X$ is equivalent to the action $G$ by left translation on the coset space $G/G_x$, for each $x \in X$.
(c) For subgroups $H$ and $K$ of $G$, the left translation actions of $G$ on $G/H$ and $G/K$ are equivalent if and only if $H$ and $K$ are conjugate.
Where I Am:
For part (a): The action of $G$ by left translation on the left coset space $G/H$ has only one orbit and is, thus, transitive. (i.e., since $gH=g \cdot H$, the orbit of $H \in G/H$ is the whole coset space.) [I pretty much understand this. It was mostly the result of piecing together definitions and properties, but doesn't make too much intuitive sense to me.]
For part (b): Since $Gx \subset X$ and we are considering transitive actions of $G$ on $X$ (i.e., actions with only one orbit), in this case, $Gx = X$ [I'm not actually sure if this is true, but it seems correct. I would like to be able to prove it, but can't seem to figure out how]. We know that there exists a bijection $f: Gx \to G/G_x$, so it must also be the case that $f$ maps from $X$ to $G/G_x$ bijectively. [Here, I need to prove that the action of $G$ on $f$ fixes $f$, which I can't figure out how to do.] Since the action of $G$ on $f$ fixes $f$, we have that $X \sim G/G_x$, as desired.
For part (c): I have made the least amount of progress on this one. To show that $H$ and $K$ are conjugate, I assume that I need to show that $K = gHg^{-1}$, for some $g \in G$, and $H = gKg^{-1}$, for some $g \in G$. I can see how this'd be the case if these subgroups were normal in $G$, but I can't seem to get it from the assumptions given. Obviously, this statement is biconditional, so I have to prove the other direction, as well; but I'd be happy to get at least One Direction$^{\text{TM}}$.