Consider the group $T_d$, the symmetries of the tetrahedron. It is isomorphic to the permutation group of four objects, so I can construct a representation of this group which is four-dimensional. I start with four basis vectors $$\left|1\right>=(1,0,0,0)$$ $$\left|2\right>=(0,1,0,0)$$ $$\left|3\right>=(0,0,1,0)$$ $$\left|4\right>=(0,0,0,1)$$ then I can construct any of the group elements in this representation according to how they permute the four vectors, e.g. $$g_{(12)(34)} = \left|1\right>\left<2\right|+\left|2\right>\left<1\right|+\left|3\right>\left<4\right|+\left|4\right>\left<3\right|$$ and there are 24 such matrices. This representation reduces to one copy of the $A_1$ trivial irrep and one copy of the $T_2$ irrep.
On the other hand, there is a three-dimensional irreducible representation (in fact the same one, $T_2$) which can be built from rotation matrices and inversions. In this representation, the matrices can be thought of as acting on these vectors $$\left|1'\right>=(1,1,1)/\sqrt{3}$$ $$\left|2'\right>=(1,1,-1)/\sqrt{3}$$ $$\left|3'\right>=(1,-1,1)/\sqrt{3}$$ $$\left|4'\right>=(-1,1,1)/\sqrt{3}$$ For example, the same group element as the one above in this representation is given by a rotation by $\pi$ about $\hat{n}=(0,0,1)$. First, I noticed that this matrix can be constructed from $$g_{(12)(34)}' = \frac{3}{4}\left(\left|1'\right>\left<2'\right|+\left|2'\right>\left<1'\right|+\left|3'\right>\left<4'\right|+\left|4'\right>\left<3'\right|\right)$$ My first question is
Why does this work even though these vectors are not orthogonal, and how is the factor of (3/4) related to the dimension of the representations? Do such relationships exist for general finite groups?
My more important and fundamental question, however, is
How can I construct a map between these two representations?
My first approach was to define $$ M = \left|1'\right>\left<1\right|+\left|2'\right>\left<2\right|+\left|3'\right>\left<3\right|+\left|4'\right>\left<4\right| $$ then I find that I can use this to map $$g_{(12)(34)}' = \frac{3}{4} M g_{(12)(34)} M^{T}$$ If I have done my work right, this looks like it will map the 4d permutation representation matrices into the 3d $T_2$ representation, but I have not yet figured out how to prove it other than by brute force. But this map doesn't appear to be invertible, because the four vectors in the $T_2$ representation are not orthogonal.
How does one generally construct maps between representations?