The group $H$ acts transitively and primitively on $\mathbb{C}=\mathbb{R}^2$. ('Primitive' means that $H$ preserves no nontrivial foliation.) It's a consequence of the classification of transitive primitive Lie transformation groups in dimension $2$ that there are only four Lie transformation groups on $\mathbb{R}^2$ that contain $H$:
- $H = \mathbb{C}\rtimes \mathbb{C}^\bullet$;
- $H\cup cH$, where $c:\mathbb{C}\to\mathbb{C}$ is conjugation: $c(z) = \bar z$;
- The orientation-preserving affine group, $\mathbb{R}^2\rtimes \operatorname{GL}^+(2,\mathbb{R})$; and
- The full affine group, $\mathbb{R}^2\rtimes \operatorname{GL}(2,\mathbb{R})$.
Of these, only $H$ itself has the property that every element is topologically conjugate to an element of $H$. (For example, every one of these groups other than $H$ contains a transformation that is not the identity but has a line of fixed points.)
Sketch of argument: Suppose that $G\supseteq H$ is a Lie group with a smooth effective left action $G\times \mathbb{R}^2\to\mathbb{R}^2$ that extends $H\times\mathbb{R}^2\to\mathbb{R}^2$, and let $G_0\subset G$ be the subgroup of $G$ that preserves the origin $O=(0,0)\in\mathbb{R}^2$. Since $H$ acts transitively on $\mathbb{R}^2$, so does $G$. Thus, $\mathbb{R}^2$ is naturally identified with $G/G_0$. Since Lie group multiplication is real-analytic in appropriate coordinates, the left action $G\times G/G_0\to G/G_0$ is also real-analytic.
Let $\mathfrak{g}$ be the Lie algebra of $G$. The action of $G$ on $\mathbb{R}^2$ induces an injective homomorphism of $\mathfrak{g}$ into the Lie algebra of real-analytic vector fields on $\mathbb{R}^2$, so one can identify $\frak{g}$ with a Lie algebra of real-analytic vector fields on $\mathbb{R}^2$. By definition, there exist real-analytic coordinates $(x,y)$ on $\mathbb{R}^2$ centered on $O$ so that the Lie algebra of $H$ is spanned by the four vector fields $X = \partial/\partial x$, $Y = \partial/\partial y$, $E = x\,\partial/\partial x + y\,\partial/\partial y$, and $R = x\,\partial/\partial y - y\,\partial/\partial x$.
For each $k\ge0$, let ${\mathfrak{g}}_k\subset \mathfrak{g}$ denote the space of vector fields in $\mathfrak{g}$ of the form $Z = f\,\partial/\partial x + g\,\partial/\partial y$ where $f$ and $g$ vanish at $O=(0,0)$ to order at least $k+1$. (Thus, $E$ and $R$ lie in ${\mathfrak{g}}_0$ but not in ${\mathfrak{g}}_1$.) Then ${\mathfrak{g}}_k\supseteq {\mathfrak{g}}_{k+1}$ and $[{\mathfrak{g}}_i,{\mathfrak{g}}_j]\subseteq {\mathfrak{g}}_{i+j}$.
If $Z$ lies in ${\mathfrak{g}}_k$ but not ${\mathfrak{g}}_{k+1}$, then at least one of $[X,Z]$ or $[Y,Z]$ does not lie in ${\mathfrak{g}}_k$. Consequently, if ${\mathfrak{g}}_k={\mathfrak{g}}_{k+1}$ for some $k$, then ${\mathfrak{g}}_j={\mathfrak{g}}_{j+1}$ for all $j\ge k$, so ${\mathfrak{g}}_k$ consists of real-analytic vector fields that vanish to infinite order at $O$, implying that ${\mathfrak{g}}_k = (0)$. Let $r\ge 1$ be the least integer for which ${\mathfrak{g}}_r = (0)$. The integer $r$ is the order of the action of $G$. (I will eventually show that $G$ has order $r=1$.)
Now,
$
\mathrm{ad}(E)(f\,\partial/\partial x {+} g\,\partial/\partial y)=[E,f\,\partial/\partial x {+} g\,\partial/\partial y]
= (Ef{-}f)\partial/\partial x + (Eg{-}g)\partial/\partial y.
$
Since $Ep = \delta p$ for a polynomial $p(x,y)$ of homogeneous degree $\delta$ in $x$ and $y$, it follows that, for $Z\in {\mathfrak{g}}_0$,
$$
(\operatorname{ad}E)(\operatorname{ad}E-1)(\operatorname{ad}E-2)\cdots\bigl(\operatorname{ad}E-(r-1)\bigr)(Z) \in {\mathfrak{g}}_r = (0).
$$
Thus, for any $Z = f\,\partial/\partial x + g\,\partial/\partial y\in {\frak{g}}_0$, the coefficients $f$ and $g$ are polynomials of degree at most $r$. Now, for any $Z\in {\mathfrak{g}}_0$, the elements $Z, \mathrm{ad}(E)(Z),\ldots, \bigl(\mathrm{ad}(E)\bigr)^r(Z)$ lie in $\mathfrak{g}_0$. It follows that, when one writes $Z = Z_1 + \dotsb + Z_r$ where $Z_i = f_i\,\partial/\partial x + g_i\,\partial/\partial y$ has $f_i$ and $g_i$ homogeneous of degree $i$, the individual terms $Z_1, \ldots, Z_r$ all lie in ${\mathfrak{g}}_0$.
Thus,
$$
{\mathfrak{g}}_0 = {\mathfrak{p}}_0 \oplus {\mathfrak{p}}_1 \oplus \dotsb \oplus {\mathfrak{p}}_{r-1}
$$
where the elements of ${\mathfrak{p}}_j$ have coefficients homogeneous of degree $j+1$ in $x$ and $y$.
The Lie algebra ${\mathfrak{p}}_0$ contains $\{E,R\}$ and lies in the span of $\{x\,\partial/\partial x,\>x\,\partial/\partial y,\> y\,\partial/\partial x,\> y\,\partial/\partial y\}$, which is a copy of ${\mathfrak{gl}}(2,\mathbb{R})$. It follows that ${\mathfrak{p}}_0$ either has dimension $2$ and is spanned by $E$ and $R$ or has dimension $4$ and is equal to ${\mathfrak{gl}}(2,\mathbb{R})$. In either case, the Lie bracket pairing ${\mathfrak{p}}_0\times {\mathfrak{p}}_1\to {\mathfrak{p}}_1$ makes ${\mathfrak{p}}_1$ into a module over the Lie algebra ${\mathfrak{p}}_0$.
If ${\mathfrak{p}}_0$ is the span of $E$ and $R$, then, since $Z\in {\mathfrak{p}}_1$ implies that both $[X,Z]$ and $[Y,Z]$ must be linear combinations of $E$ and $R$, it's easy to check that such a $Z$ has to be a linear combination of $A = (x^2{-}y^2)\,\partial/\partial x + 2xy\,\partial/\partial y$ and $B = 2xy\,\partial/\partial x - (x^2{-}y^2)\,\partial/\partial y$.
Moreover, if ${\mathfrak{p}}_1\not=(0)$, then it must be of dimension $2$ and spanned by these two vector fields. However, the flow of $A$ is clearly not complete (just look at its action on the line $y=0$), and all of the vector fields in $\mathfrak{g}$ must be complete. Hence ${\mathfrak{p}}_1=(0)$, and $G$ has order $1$ and the identity component of $G$ must equal $H$.
If ${\mathfrak{p}}_0 = {\mathfrak{gl}}(2,\mathbb{R})$, then ${\mathfrak{p}}_1$ must be a sum of irreducible ${\mathfrak{p}}_0$ modules. It is easy to see that the span of $xE$ and $yE$ is an irreducible ${\mathfrak{gl}}(2,\mathbb{R})$-module of dimension $2$, while the space of vector fields of the form $f_y\,\partial/\partial x - f_x\,\partial/\partial y$, where $f(x,y)$ is a homogeneous cubic polynomial, is an irreducible ${\mathfrak{gl}}(2,\mathbb{R})$-module of dimension $4$. Together these two spaces span all of the vector fields with quadratic coefficients. However, the flow of the vector field $xE$ is not complete, and the flow of the vector field $f_y\,\partial/\partial x - f_x\,\partial/\partial y$ with $f = x^2y$ is not complete.
Thus, ${\mathfrak{p}}_1=(0)$ in this case as well. Thus, the action of $G$ has order $1$, and, for dimension reasons, the identity component of $G$ must be the orientation-preserving affine group $\mathbb{R}^2\rtimes \mathrm{GL}^+(2,\mathbb{R})$.
Now, in either case, the abelian subalgebra $\mathfrak{s}$ of $\mathfrak{g}$ spanned by $X$ and $Y$ is easily seen to equal the null space of the Killing form of $\mathfrak{g}$, and, hence, it is preserved by any automorphism of $\mathfrak{g}$. In particular, the subgroup $S\subset G$ of translations in $\mathbb{R}^2$ is preserved by any automorphism of the identity component of $G$.
Let $\ell\in G_0$ be any element of $G$ that fixes $O\in\mathbb{R}^2$. Since $\ell S \ell^{-1} = S$, it follows that $\ell:\mathbb{R}^2\to\mathbb{R}^2$ must be a linear map. Moreover, it must normalize the identity component of the subgroup $G_0\subset\mathrm{GL}(2,\mathbb{R})$, which is either $C^\bullet$ or $\mathrm{GL}^+(2,\mathbb{R})$. In the first case, this implies that $G=H$ or $H\cup cH$, and in the second case this implies that $G = \mathbb{R}^2\rtimes\mathrm{GL}^+(2,\mathbb{R})$ or $G = \mathbb{R}^2\rtimes\mathrm{GL}(2,\mathbb{R})$.
