Starting with an observation: If $A$ were the center of $c$, then $H$ would be the pole of line $FG$ with respect to that circle. In that case, the fact that conics map to conics could be easily seen in an analytic description from the fact that a polar-to-pole transformation is only a matrix multiplication.
Maybe that is something we can show in general? I'm afraid my way of doing so has a lot of analytic brute-force aspects to it, so it might not be exactly what you were hoping for, but perhaps it can still be of some use.
Let's start by assuming, without loss of generality, that $c$ is the unit circle, i.e. in homogeneous coordinates $x^2+y^2-z^2=0$. Let's denote your line $FG$ as $g$, and pick generic homogeneous coordinates for it and for the point $A$:
$$
A=\begin{bmatrix}A_1\\A_2\\A_3\end{bmatrix}\qquad
g=\begin{bmatrix}g_1\\g_2\\g_3\end{bmatrix}
$$
The points where $g$ intersects $c$ I will call $G_{1,2}$; they correspond to your $F,G$.
$$
G_{1,2}=\begin{bmatrix}-g_1g_3\\-g_2g_3\\g_1^2+g_2^2\end{bmatrix}\pm
\lambda\begin{bmatrix}g_2\\-g_1\\0\end{bmatrix}\quad
\text{with }\lambda:=\sqrt{g_1^2+g_2^2-g_3^2}
$$
Now we need to construct $H$ from these $G_i$. Start by connecting $G_i$ to $A$ using a cross product. Take the point at infinity in the orthogonal direction by multiplying with
$$J:=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}$$
Connect that point at infinity with $G_i$ again to get the perpendicular line in that point. And last intersect the two lines you obtained in this fashion. In formula:
\begin{align*}
h_i&=J(A\times G_i)\times G_i \\
H&=h_1\times h_2
\end{align*}
If you feed this to a computer algebra system of your choice, the result may look massive at first glance. But there is a lot of common factors in there. Specifically I was able to identify a common factor of
$$\mu:=2(g_1^2+g_2^2)^3\sqrt{g_1^2+g_2^2-g_3^2}$$
The fact that this term must not be zero can be seen as a non-degeneracy constraint, requiring your line to be neither a tangent to $c$ nor the line at infinity (or some other line with complex coordinates passing through one of the ideal circle points). You can cancel that factor (and remove the non-degeneracy constraint) since we are dealing with homogeneous coordinates. The result can then be expressed as
$$H=\mu Mg\sim Mg\quad\text{with }M:=\begin{bmatrix}
A_2^2-A_3^2 & -A_1A_2 & -A_1A_3 \\
-A_1A_2 & A_1^2-A_3^2 & -A_2A_3 \\
A_1A_3 & A_2A_3 & A_3^2
\end{bmatrix}$$
Remember that the above is w.l.o.g. assuming $c$ to be the unit circle. For other circles, the entries of $M$ would need to take your choice of circle into account.
So we can confirm that the map from $g$ to $H$ is a simple matrix multiplication. For the special case discussed initially, where $A$ is the center of the unit circle $c$ you get $A_1=A_2=0$ and $M$ becomes the matrix of the unit circle, which is exactly what you multiply with to convert polar to pole.
You can obtain $g$ from $K$ by multiplying with the primal matrix of $d$, so this shows that the map $K\to H$ is just a matrix multiplication as well and as such will preserve conic cross ratios, concluding one of your attempts for a solution.
And to be extra clear, here is how $M$ helps you to get one conic from the other. If you write $D_d$ as the dual matrix of conic $d$ (i.e. a line is tangent if the quadratic form becomes zero), and $P_e$ as the primal matrix of conic $e$ (i.e. a point is incident if the quadratic form becomes zero) then up to a scalar factor you get
$$
H^TP_eH = (Mg)^TP_e(Mg) = g^T(M^TP_eM)g = g^TD_dg \\
D_d = M^TP_eM\qquad P_e = (M^T)^{-1}D_dM^{-1}
$$
For non-degenerate conics you can use matrix inversion to convert between primal and dual representation of each conic.
Note that if $A_1^2+A_2^2-A_3^2=0$ or $A_3=0$ then $\det(M)=0$ so you would get a singular matrix which maps a non-degenerate conic $d$ to a degenerate conic $e$ in a way that can not be reversed. You might want to require $A$ to neither lie on $c$ nor on the line at infinity if you want to avoid this situation.
