Transforming between matrices of SO($3$) The $SO(3)$ $3\times 3$ matrices are well known in physics:
$J_1=\frac{1}{\sqrt{2}}\left(\matrix{0&1&0\\1&0&1\\ 0&1&0}\right) \ \ ; \ \ J_2=\frac{1}{\sqrt{2}}\left(\matrix{0&-i&0\\i&0&-i\\ 0&i&0}\right) \ \ ; \ \ J_3=\left(\matrix{1&0&0\\0&0&0\\ 0&0&-1}\right)$
However, we can also build another group of 3x3 matrices that still satisfy the same Lie algebra $so(3)$:
$K_1=\left(\matrix{0&0&0\\0&0&-i\\ 0&i&0}\right) \ \ ; \ \ K_2=\left(\matrix{0&0&i\\0&0&0\\ -i&0&0}\right) \ \ ; \ \ K_3=\left(\matrix{0&-i&0\\i&0&0\\ 0&0&0}\right)$
Which I found in the 5th page of this document https://www.uio.no/studier/emner/matnat/fys/FYS3120/v12/grpt3120E.pdf.
Since the representation of $so(3)$ is unique these matrices should be related to one another, particularly by a transformation of the form:
$K_i=SJ_iS^{-1}$
Where $S$ is a $3\times 3$ unitary matrix. While I know how to convert one couple of matrices from one to another (for example, an $S_1$ that converts only $K_1\rightarrow J_1$), I've never seen how to convert from one group of matrices to another. I've looked everywhere and tried different combinations, but I can't seem to find a single matrix $M$ that transforms all the $K_i$ to $J_i$.
I just found in an article that the matrix $S$ should have the following form,
$$S=\frac{1}{\sqrt{2}}\left(\matrix{-1&0&1\\ -i&0&-i\\ 0&\sqrt{2}&0} \right).$$
However, I'm still trying to see where it came from. Could you give me any guidance, or if there's a general procedure?
Edit: I found out that this process is changing a basis of orthonormal matrices from one basis to another. I still haven't found any explicit procedure, though.
 A: Related. 
The idea is to transform the 3d vectors these sets of matrices act on.
Consider the eigenvectors of $K_3$, the transposes of $(-1,-i,0)/\sqrt{2}$ for eigenvalue 1; of $(0,0,1)$ for null eigenvalue; and of $(1,-i,0)/\sqrt{2}$ for eigenvalue -1. Utilize them as column vectors of your matrix S.
What basis change converts them to the evident eigenvectors of $J_3$, the defining structure of the spherical basis? S is then the diagonalizing transformation for it, namely $S^{-1} K_3 S\equiv J_3$.
Having found this basis change, how does it transform all 3 matrices from the Cartesian to the spherical basis? You are done. That is, the other two generators come in for the ride: The $J_2, J_3$ were found this way. They are not arbitrary matrices: they are defined as similarity transforms of the Ks preserving the Lie algebra.


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*The point is all 3 generators, in any basis, have to obey the same Lie algebra, so they must transform under the very same similarity transformation to preserve this algebra.

