Name the corners of a square as 1,2,3,4 in clockwise order.
As you know: The group of all rigid motions of the square back to itself, called $D_4$, has eight elements, written in cycle form as: $\begin{align*} \mathrm{id},& &\qquad a&=(1234),\\ b&=(13)(24), &c&=(1432),\\ d&=(12)(34), &e&=(14)(23),\\ f&=(13), &g&=(24). \end{align*}$
The two subgroups $x_1= \langle d, e\rangle$ and $x_2=\langle f,g\rangle$ are each "abstractly" isomorphic to the Klein 4-group, and share the central element $b$.
But $x_1$ and $x_2$ are not conjugate as subgroups of $D_4$ (or as subgroups of $S_4$, since they have different cycle structure, and half of $x_2$ is odd while all of $x_1$ is even).
$D_4$s regular representation in $A_8$ can be written as $\begin{align*} \mathrm{id}, &&\quad A&=(1234)(5768);\\ B&=(13)(24)(56)(78);& C&=(1432)(5867);\\ D&=(15)(28)(36)(47);& E&=(16)(27)(35)(48);\\ F&=(17)(25)(38)(46);& G&=(18)(26)(37)(45). \end{align*}$ The image of $x_1$ is $\langle D,E\rangle$ and that of $x_2$ is $\langle F,G\rangle$.
But, $\langle D,E\rangle$ is automorphic to $\langle F,G\rangle$ under the conjugation in $S_8$ by $(5768)$.
My question: Is there a way to draw $D_4$ as motions of a geometric figure that would allow me to see the automorphism?
Thanks,