Note that the common diagonal is a diameter of the circle that will ultimately circumscribe the collection of rectangles. Each rectangle is composed of four chords of the circle, and the set of outermost chords will form a regular $n$-gon if a vertex lands on $B$. Because each new card contributes two sides of the polygon, the problem is asking for $\frac{n}{2}$. Let $O$ be the center of the circle.

Let $\theta = \angle ACB$. Then the measure of minor arc $\widehat{AB}$ is $2\theta$, and $n = \frac{360^\circ}{2\theta}$. Because $AB = 1$ and $BC = 2 + \sqrt{3}$,
$$\tan \theta = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}.$$
Each card is rotated through an angle $\angle AOB = 2\theta$ compared to the previous card. A Double Angle Formula gives
$$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{2(2 - \sqrt{3})}{1 - (2 - \sqrt{3})^2} = \frac{1}{\sqrt{3}}.$$
This value is recognizable as $\tan 30^\circ$, so $2\theta = 30^\circ$ and $n = \frac{360^\circ}{30} = 12$. The polygon is a dodecagon. It will take just $\frac{n}{2} = 6$ cards to complete the dodecagon and have a new vertex land on vertex $B$, as illustrated below.