Problem
Assuming an ellipse E defined by $(a, b)$ respectively its semi-minor and semi-major axes.
We define $R$ the rectangle of height $a$ and of length $b$, with its sides tangent to the ellipse on the vertices (and co-vertices).
We define $D_n$ a rhombus in which the ellipse can be inscribed, with $n \in [0, 90]$ the angle between a semi-minor axis and the ray from the center of the ellipse to the point in which the rhombus is tangential to the ellipse, such that $\forall n \in [0, 90], R \neq D_n$.
We know that in the extreme cases ($a=0$, $b=0$, ...) $D_n$ isn't well defined for it is equal to $R$.
But in the general case, do we know which shape (rectangle or rhombus) is the smallest? What are the formulas?
Schema
Here is a link to a Geogebra sheet illustrating the problem.
If by "smallest" you mean the rhombus $D_n$ that has minimal area, then this is easily solved as follows.
Recall that area is proportional under scaling transformations. So if we perform the transformation that maps the ellipse to the unit circle, namely $$(x', y') = (x/a, y/b),$$ then the line tangent to a point on the unit circle at $(x', y') = (\cos \theta, \sin \theta)$ for $0 < \theta < \pi/2$ encloses an area in the first quadrant of $$A'(\theta) = \frac{1}{2} \sec \theta \csc \theta = \csc 2\theta.$$ Hence the area after the inverse transformation is $$A(\theta) = ab \csc 2\theta.$$ This area is minimized for a fixed $a, b$ when $\csc 2\theta$ is minimized, which of course occurs when $\theta = \pi/4$. But this is the angle measure in the transformed coordinate system, which in general does not preserve angles. In the original coordinate system, as you have defined $n$, we must have $$\tan n = \frac{a}{b} \tan \frac{\pi}{4} = \frac{a}{b},$$ hence $$n = \tan^{-1} \frac{a}{b}.$$