On paper we, orthogonal, project a cube as in provided image. But is it always really a cube? And, in this particular example, the 8 different 2D coordinates of the vertices have integer values. But is that even possible? I believe it is not possible (integer valued coordinates for distinct projections), but, is there a, perhaps simple, proof of this general impossibility, or, is there an example of a possibility?
Possible with all $8$ vertices having different integer $(x,y)$ coordinates, for example:
$$\require{cancel} \begin{array}{cccc} & (-9,20,\cancel{12}) & ----- & (11,20,\cancel{27})\\ (-21,5,\cancel{28}) & ----- & (-1,5,\cancel{43}) & \mid\\ \mid & \vdots & \mid & \mid\\ \mid & \vdots & \mid & \mid\\ \mid & \vdots & \mid & \mid\\ \mid & \vdots & \mid & \mid\\ \mid &(0,0,\cancel0) & \cdots\cdots\mid \cdots\cdots& (20,0,\cancel{15})\\ (-12,-15,\cancel{16}) & ----- & (8,-15,\cancel{31})\\ \end{array}$$
This example is based on the $3-4-5$ right-angled triangle, by rotating a cube twice along two axes by $\tan^{-1}(3/4)$. The cube side length is $25$.
To create this type of example, first start with a unit cube with vertices $\mathbf v\in\{0,1\}^3$.
Pick a Pythagorean triple where $a^2+b^2=c^2$, then scale each vertex by $c$ and rotate about the $x$-axis:
$$\mathbf v \mapsto c\begin{pmatrix} 1 & 0 & 0\\ 0 & b/c & -a/c\\ 0 & a/c & b/c \end{pmatrix}\mathbf v = \begin{pmatrix} c & 0 & 0\\ 0 & b & -a\\ 0 & a & b \end{pmatrix}\mathbf v$$
Then pick another (or the same) Pythagorean triple where $d^2+e^2=f^2$, then scale the above result by $f$ and rotate about the $y$-axis:
$$\begin{align*} T(\mathbf v) &= f\begin{pmatrix} e/f & 0 & -d/f\\ 0 & 1 & 0\\ d/f & 0 & e/f \end{pmatrix}\begin{pmatrix} c & 0 & 0\\ 0 & b & -a\\ 0 & a & b \end{pmatrix}\mathbf v\\ &= \begin{pmatrix} e & 0 & -d\\ 0 & f & 0\\ d & 0 & e \end{pmatrix}\begin{pmatrix} c & 0 & 0\\ 0 & b & -a\\ 0 & a & b \end{pmatrix}\mathbf v\\ &= \begin{pmatrix} ce & -ad & -bd\\ 0 & bf & -af\\ cd & ae & be \end{pmatrix}\mathbf v\\ \end{align*}$$
Lastly, project the above result parallel to the $z$-axis, and only keep the $(x,y)$-coordinates unchanged:
$$\begin{align*} P(T(\mathbf v)) &= \begin{pmatrix} 1 & 0&0\\ 0 & 1 & 0 \end{pmatrix}T(\mathbf v)\\ &= \begin{pmatrix} ce & -ad & -bd\\ 0 & bf & -af \end{pmatrix}\mathbf v\\ \end{align*}$$
The example above is from picking $3^2+4^2=5^2$ twice.
For your initial projection, by hand calculation, for now I can only fit a $2\sqrt{10}\times 2\sqrt{10}\times 4\sqrt{3}$ cuboid, i.e. a $\sqrt5 :\sqrt 5 :\sqrt 6$ cuboid.