You will always get the set of all points outsude of the inner shape exactly distance 1 from the inner shape. For a circle or ellipse, you get anityer circle or ellipse with the radius or semiaxes increased in length by 1.
For a square, you add a rectangle of thickness 1 to each edge and then add quarter circles at the corner to fill in the gaps.
By the way, have you heard of extremal length of an annulus?https://en.m.wikipedia.org/wiki/Extremal_length
It is really cool. It assigns a number to each ring that tells how 'thick' it is using complex analysis. It is hard to compute but useful.
I assume what you want to do is simulate the effect of taking your 2D shape, putting it into 3D space, rotating it somehow, then projecting it orthogonally back into its original 2D plane.
The general idea is to pick your favourite 3D rotation, represented as a $3\times3$ linear transformation matrix, then cut out the top $2\times2$ submatrix and use that as a 2D linear transformation.
Example: Rotation around the Z axis by angle $\alpha$.
The $3\times3$ matrix corresponding to this rotation looks like this:
$$\begin{pmatrix}\cos\alpha&-\sin\alpha&0\\\sin\alpha&\cos\alpha&0\\0&0&1\end{pmatrix}$$
The top left $2\times2$ submatrix is $\begin{pmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{pmatrix}$, which corresponds to a 2D rotation by angle $\alpha$.
Example: Rotation around the Y axis by angle $\alpha$.
The $3\times3$ matrix corresponding to this rotation looks like this:
$$\begin{pmatrix}\cos\alpha&0&-\sin\alpha\\0&1&0\\\sin\alpha&0&\cos\alpha\end{pmatrix}$$
The top left $2\times 2$ submatrix is $\begin{pmatrix}\cos\alpha&0\\0&1\end{pmatrix}$, which corresponds to an X-axis scaling by $\cos\alpha$.
Case you might be interested in: Rotation around the X axis by $\alpha$, then around the Y axis by $\beta$.
The $3\times3$ matrix corresponding to this rotation looks like this:
$$\begin{pmatrix}\cos\beta&0&-\sin\beta\\0&1&0\\\sin\beta&0&\cos\beta\end{pmatrix} \times \begin{pmatrix}1&0&0\\0&\cos\alpha&-\sin\alpha\\0&\sin\alpha&\cos\alpha\end{pmatrix} = \begin{pmatrix}\cos\beta&-\sin\alpha \sin\beta&-\cos\alpha \sin\beta\\
0&\cos\alpha&-\sin\alpha\\
\sin\beta&\cos\beta \sin\alpha&\cos\alpha \cos\beta\end{pmatrix}.$$
The top left $2\times 2$ submatrix is $\begin{pmatrix}\cos\beta&-\sin\alpha\sin\beta\\0&\cos\alpha\end{pmatrix}$. This can be decomposed into a series of axis-aligned scalings and skewings like this:
$$\begin{pmatrix}\cos\beta&-\sin\alpha\sin\beta\\0&\cos\alpha\end{pmatrix} = \begin{pmatrix}1&0\\0&\cos\alpha\end{pmatrix} \times \begin{pmatrix}1&-\sin\alpha\sin\beta\\0&1\end{pmatrix} \times \begin{pmatrix}\cos\beta&0\\0&1\end{pmatrix}$$
So you can simulate the effect by scaling the shape in X direction by a factor of $\cos\beta$, then skewing it parallel to the X axis by a factor of $-\sin\alpha\sin\beta$, then scaling the shape in Y direction by a factor of $\cos\alpha$.
Of course, if you have the relative corner coordinates available, then it's probably simpler to just use the entries of the 2x2 matrix directly to transform them:
\begin{align}x_{new} &= \cos\beta\cdot x_{old} -\sin\alpha\cdot\sin\beta\cdot y_{old}\\y_{new} &= \cos\alpha\cdot y_{old}\end{align}
Best Answer
You can follow your anchor point (projection) approach using a point in the interior. That will preserve angles. It is not clear how you got from A to B in the second drawing. The lines between corresponding vertices do not meet in a point, which is why the angles change.