Problem with classes of conjugation Let $H$ be any non-trivial group with neutral element $e$. Let $\circ$ be an action of a group $G:=H\times H$ on the set $X:=H$ given by $(g,h)\circ x:=gxh^{-1}$.

*

*Prove that an action $\circ$ is transitive,

*Prove that if an action $\circ$ is $2$-transitive then $H$ has got exactly one non-trivial class of conjugation (that is: $H=a^H\overset{\cdot}{\cup}e$ for certain $a \in H$, where $\overset{\cdot}{\cup}$ indicates the disjoint sum),

*Prove that if $H$ has got exactly one non-trivial class of conjugation, then an action $\circ$ is $2$-transitive.

My attempt
I managed only to do the first part. Taking $g:=x \in H$ and $h:=x \in H$ we obtain $(x,x)\circ x:=xxx^{-1}=xe=x.$ Is it proper approach?
Unfortunately I have big problem with $2.$ and $3.$.
Thanks.

Reminder
Let me remind you that the action group $G$ on the set $X$ we name $2$-transitive if for every $x_1$, $x_2$, $x_1'$, $x_2'\in X$ s. t. $x_1\neq x_2$ and $x_1'\neq x_2'$, there exists $g \in G$, s. t. $gx_1=x_1'$ and $gx_2=x_2'$. What is more class of conjugation of  group's $H$ element $a$ we name the set $a^H:=\{hah^{-1}:h\in H\}$.
 A: Point 1.
The action is transitive if for every $h_1,h_2\in H$ (= the acted on set, in your case), there exists some $(h,h')\in H\times H$ (= the acting group, in your case) such that $(h,h')\circ h_1=h_2$; now, this latter holds for, e.g., $(h,h')=(h_2,h_1)$ (but also $(h,h')=(h_1^{-1},h_2^{-1})$ would fit, etc.), so the action is transitive.
[ Alternatively, $\operatorname{Stab}(h)=\{(h',h'')\in H\times H\mid h'hh''^{-1}=h\}$; now, given $h\in H$, $\forall h''\in H, \exists h'''\in H\mid h''=h'''h$, and hence:
\begin{alignat}{1}
\operatorname{Stab}(h) &= \{(h',h''')\in H\times H\mid h'hh^{-1}h'''^{-1}=h\} \\
&= \{(h',h''')\in H\times H\mid h'h'''^{-1}=h\} \\
&= \{(h',h^{-1}h')\in H\times H\mid h'\in H\} \\
\end{alignat}
whence $|\operatorname{Stab}(h)|=|H|$ for every $h\in H$. By the Orbit-Stabilizer Theorem, $|O(h)|=\frac{|H\times H|}{|\operatorname{Stab}(h)|}=\frac{|H|^2}{|H|}=|H|$ for every $h\in H$, so there is one orbit only, and the action is transitive. ]

Point 2.
The action "$\circ$" is $2$-transitive if and only if (definition) for every $(h_1,h_2),(\tilde h_1,\tilde h_2)\in H\times H$, with $h_1\ne h_2$ and $\tilde h_1\ne\tilde h_2$, there is some $(h,h')\in H\times H$ such that $\tilde h_1=hh_1h'^{-1}$ and $\tilde h_2=hh_2h'^{-1}$. From this latter we get $h=\tilde h_2h'h_2^{-1}$, which replaced in the former gets $\tilde h_1=\tilde h_2h'h_2^{-1}h_1h'^{-1}$, whence $\tilde h_2^{-1}\tilde h_1=h'h_2^{-1}h_1h'^{-1}$. By the arbitrariness of $h_1,\tilde h_1, h_2,\tilde h_2$ (with the stated inequalities), this latter means that all the non-identity elements of $H$ are mutually conjugated, so there is just one nontrivial conjugacy class.
