Since I refered to this thread while answering another, I might as well answer this one, albeit being more than two years late. First, yes, if $X$ is a left $G$-set, then $G$ acts on $X$ transitively if and only if $G\cdot x=X$ for at least one (whence for any) $x\in X$.
For Part (a), you got it correctly. For Part (b), for a fixed $x\in X$, we define a function $f:X\to G/G_x$ as $f(y)=gG_x$, if $y\in X$ and $g\in G$ satisfy $y=g\cdot x$ (since $G$ acts on $X$ transitively, such an element $g\in G$ exists for every $y\in X$). Here, $G_x$ is the stabilizer of $x$, namely $G_x:=\big\{g\in G\,\big|\,g\cdot x=x\big\}$. We need to prove that $f$ is well defined. Suppose that $y=g\cdot x$ and $y=h\cdot x$ for $g,h\in G$. We have to prove that $gG_x=hG_x$. This is easy, since $$(h^{-1}g)\cdot x=h^{-1}\cdot(g\cdot x)=h^{-1}\cdot y=h^{-1}\cdot(h\cdot x)=(h^{-1}h)\cdot x=1_G\cdot x=x\,,$$
whence $h^{-1}g\in G_x$, and so $gG_x=hG_x$. Injectivity of $f$ is clear: if $f(y)=f(z)$ for some $y=g\cdot x$ and $z=h\cdot x$ with $g,h\in G$, then $g G_x=h G_x$, or $h^{-1}g\in G_x$, whence
$$y=g\cdot x=(hh^{-1}g)\cdot x=h\cdot\big((h^{-1}g)\cdot x\big)=h\cdot x=z\,.$$ Surjectivity of $f$ is also immediate: for $g\in G$, we have $f(g\cdot x)=g G_x$.
For Part (c), we assume first that there exists an isomorphism of left $G$-sets $\phi:G/H\to G/K$. Suppose that $\phi(H)=kK$ for some $k\in G$. We know that $h\in H$ implies that $hH=H$, so
$$kK=\phi(H)=\phi(hH)=\phi(h\cdot H)=h\cdot\phi(H)=hkK\,.$$
This proves that $k^{-1}hk\in K$ for every $h\in H$. That is, $k^{-1}Hk\subseteq K$. On the other hand,
$$\phi(k^{-1}H)=\phi(k^{-1}\cdot H)=k^{-1}\cdot\phi(H)=k^{-1}\cdot (kK)=K\,.$$
If $t\in K$, then $tK=K$, so
$$\phi(k^{-1}H)=K=tK=t\cdot K=t\cdot \phi(k^{-1}H)=\phi\big(t\cdot (k^{-1}H\big)=\phi(tk^{-1}H)\,.$$
Since $\phi$ is injective, $k^{-1}H=tk^{-1}H$, or $H=ktk^{-1}H$, which means $kKk^{-1}\subseteq H$, or equivalently $K\subseteq k^{-1}Hk$. That is, $K=k^{-1}Hk$ is a conjugate of $H$. Conversely, if $K=k^{-1}Hk$ for some $k\in G$, then we can define an isomorphism of left $G$-sets $\phi:G/H\to G/K$ by sending $gH\mapsto gkK$ for all $g\in G$. It is not difficult to show that $\phi$ is well defined, injective, surjective, and compatible with the left $G$-actions on $G/H$ and on $G/K$.
In addition, if $X$ and $Y$ are transitive left $G$-sets, then $X$ and $Y$ are isomorphic as left $G$-sets if and only if, for some $x\in X$ and $y\in Y$, the stabilizing subgroups $G_x$ and $G_y$ are conjugate subgroups of $G$. This follows immediately from Part (b) and Part (c) of this problem.
P.S. To avoid possible stupid problems with the empty set, all sets that appear in this answer are inherently assumed to be nonempty.
Best Answer
You have proved that $G_x=\{e\}$. But, by definition, $G_x=\{g\in G\mid gxg^{-1}=x\}$, whence $x\in G_x$ (in fact, $xxx^{-1}=x$). Therefore, $x=e$. But this all holds for every $x\in G$, and hence $G=\{e\}$. Namely, for nontrivial $G$, the action of $G$ on itself by conjugation is never transitive.