What formulas are required to calculate a 3d transformation? Considering the point of view of square A and B, what math tranformations must be applied (either to the 3d camera or world) to transition from A to B?
I can tell that for the B viewpoint I had to move right and up, but I lack the math background to know what formulas can give me accurate values. Thanks

 A: A way to do it is to rotate the cube around the $z$ axis, then around the $x$ axis, with a $\pi/4$ rotation angle.
It means that if we define classically the rotation matrices around these resp. axes as:
$$R_z=\begin{pmatrix}\cos(a)& -\sin(a)& 0\\
   \sin(a)& \cos(a)& 0\\
   0& 0& 1\end{pmatrix}
R_x=\begin{pmatrix}1& 0& 0\\
   0& \cos(b)& -\sin(b)\\
   0& \sin(b)& \cos(b)\end{pmatrix},$$
with $a=b=\pi/4$, the rotation to be applied to the cube is the following product:
$$R=R_x R_z$$
(Recall: matrices must be applied from right to left), giving:
$$R \approx \begin{pmatrix}0.7071  & -0.7071  &     0\\
    0.5000  &  0.5000 &  -0.7071\\
    0.5000  &  0.5000 &   0.7071 \end{pmatrix}$$
But, if we keep $a=\pi/4$ and take different values of $b$, it will make your cube do a kind of curtsy in front of you, helping you to select the most convenient $b$.
A particular case: $b=\operatorname{acos}(1/\sqrt{3})$, gives a position with 3 identical lozenges:
$$R \approx \begin{pmatrix}0.7071  & -0.7071  &       0\\
    0.4082  &  0.4082 &  -0.8165\\
    0.5774  &  0.5774 &    \ \ 0.5774\end{pmatrix}$$
A: I would recommend an approach where you first determine the initial basis unit vectors $\hat{u}$, $\hat{v}$, $\hat{w}$ and a fixed point $\vec{t}$, and corresponding final basis unit vectors $\hat{U}$, $\hat{V}$, $\hat{W}$ and a fixed point $\vec{T}$, such that the basis vectors are orthonormal,
$$\begin{array}{rcr}
\hat{u} \cdot \hat{u} = 1 & ~ & \hat{U} \cdot \hat{U} = 1 \\
\hat{u} \cdot \hat{v} = 0 & ~ & \hat{U} \cdot \hat{V} = 0 \\
\hat{u} \cdot \hat{w} = 0 & ~ & \hat{U} \cdot \hat{W} = 0 \\
\hat{v} \cdot \hat{u} = 0 & ~ & \hat{V} \cdot \hat{U} = 0 \\
\hat{v} \cdot \hat{v} = 1 & ~ & \hat{V} \cdot \hat{V} = 1 \\
\hat{v} \cdot \hat{w} = 0 & ~ & \hat{V} \cdot \hat{W} = 0 \\
\hat{w} \cdot \hat{u} = 0 & ~ & \hat{W} \cdot \hat{U} = 0 \\
\hat{w} \cdot \hat{v} = 0 & ~ & \hat{W} \cdot \hat{V} = 0 \\
\hat{w} \cdot \hat{w} = 1 & ~ & \hat{W} \cdot \hat{W} = 1 \\
\hat{u} \times \hat{v} = \hat{w} & ~ & \hat{U} \times \hat{V} = \hat{W} \\
\hat{w} \times \hat{u} = \hat{v} & ~ & \hat{W} \times \hat{U} = \hat{V} \\
\hat{v} \times \hat{w} = \hat{u} & ~ & \hat{V} \times \hat{W} = \hat{U} \\
\end{array}$$
Next, form two $4 \times 4$ transformation matrices:
$$\mathbf{M}_\text{before} = \left[ \begin{matrix}
u_x & v_x & w_x & t_x \\
u_y & v_y & w_y & t_y \\
u_z & v_z & w_z & t_z \\
  0 &   0 &   0 &   1 \\
\end{matrix} \right ], \quad \mathbf{M}_\text{after} = \left[ \begin{matrix}
U_x & V_x & W_x & T_x \\
U_y & V_y & W_y & T_y \\
U_z & V_z & W_z & T_z \\
  0 &   0 &   0 &   1 \\
\end{matrix} \right ]$$
Then, you can calculate the transformation matrix $\mathbf{M}$ using
$$\mathbf{M} = \mathbf{M}_\text{after} \mathbf{M}_\text{before}^{-1}$$
noting that because of orthonormality,
$$\mathbf{M}_\text{before}^{-1} = \left[ \begin{matrix}
v_y w_z - v_z w_y & v_z w_x - v_x w_z & v_x w_y - v_y w_x & t_x (v_z w_y - v_y w_z) + t_y (v_x w_z - v_z w_x) + t_z (v_y w_x - v_x w_y) \\
u_z w_y - u_y w_z & u_x w_z - u_z w_x & u_y w_x - u_x w_y & t_x (u_y w_z - u_z w_y) + t_y (u_z w_x - u_x w_z) + t_z (u_x w_y - u_y w_x) \\
u_y v_z - u_z v_y & u_z v_x - u_x v_z & u_x v_y - u_y v_x & t_x (u_z v_y - u_y v_z) + t_y (u_x v_z - u_z v_x) + t_z (u_y v_x - u_x v_y) \\
0 & 0 & 0 & 1 \\
\end{matrix} \right]$$
but this also applies to non-orthonormal basis vector sets, as long as the initial vector $\vec{u}$ corresponds to final vector $\vec{U}$, initial vector $\vec{v}$ to final $\vec{V}$, initial $\vec{w}$ to final $\vec{W}$, they are all nonzero, and initial vector $\vec{t}$ corresponds to final vector $\vec{T}$.
The resulting transformation matrix $\mathbf{M}$ is applied to point $\vec{p} = ( x, y, z )$, transforming it to $\vec{q} = ( \chi , \gamma, \zeta )$, via
$$\left[\begin{matrix} \chi \\ \gamma \\ \zeta \\ 1 \end{matrix} \right] = \mathbf{M} \left[ \begin{matrix} x \\ y \\ z \\ 1 \end{matrix} \right]$$
and represents both rotation and translation: the upper left $3\times 3$ matrix of $\mathbf{M}$ is a rotation matrix, and upper right $3 \times 1$ vector is the translation after rotation.
These $4\times 4$ matrices are extremely common in 3D graphics, in e.g. OpenGL.
