Proving $AD_1A^{-1}=D_2$ I want to prove that if $A$ is a permutation matrix, and $D_1$ is diagonal, than $AD_1A^{-1}=D_2$ where $D_2$ is also a diagonal matrix.
I have worked out that $A^{-1}=A^T$ and I can see that the positions of $A$ are the same positions $AD_1$ has filled, but in $A$ they are $1$'s obviously and in $AD_1$ they are the elements of $D_1$ rearranged(by row I believe). Then $A^T$ is going to rearrange columns to the diagonal form, but I don't know how to prove this. Thanks.
 A: (1) Consider the map $$ \rho : S_n \rightarrow M_n(\mathbb{R}^n) $$ 
where $S_n$ is symmetric group. So if $A$ is a permutation, then it is an image of $\rho$. Let $\rho(t)=A$. And note that if $t=t_1\cdots t_m$ where $t_i$ is a single permutation, then $$ A_i:=\rho(t_i),\ A=A_1\cdots A_m$$ 
To show that $ADA$ where $D$ is diagonal, we suffice to show that $A_iDA_i$ is diagonal.
(2) Hence we can assume that $A$ is corresponded to $t=(i,j)$, simply $(12)$. If $D$ is diagonal, then $ADA$ is diagonal when we compute directly. 
A: If $A$ is the permutation matrix corresponding to the permutation $\pi$, and $B=(b_{ij})$, then $AB=(b_{\pi(i)j})$ and $BA^T=(b_{i\pi(j)})$, thus $ABA^T=(b_{\pi(i)\pi(j)})$.
A diagonal matrix $D$ is defined by $i\ne j\implies d_{ij}=0$.
Assume $i\ne j$.
We have for the elements of $D'$: $d'_{ij} = d_{\pi^{-1}(i)\pi^{-1}(j)}$
Since a permutation is a bijection, we have $\pi(i)=\pi(j)\iff i=j$, and thus for $i\ne j$, we have $\pi^{-1}(i)\ne\pi^{-1}(j)$.
But since $D$ is diagonal, this means that $d'_{ij} = d_{\pi^{-1}(i)\pi^{-1}(j)}=0$. That is, $D'$ is diagonal.
A: This is one of those many occasions where a more abstract approach will facilitate understanding.
What does is mean that a linear operator $\phi$ has a diagonal matrix with respect to a given basis$\def\B{\mathcal B}~\B$? Answer: that every vector of$~\B$ is an eigenvector of$~\phi$. So the diagonal matrix $D_1$ represents a linear operator on$~K^n$ (with $K$ your field) for which all the standard basis vectors are eigenvectors.
What does the matrix $AD_1A^{-1}$ represent? Answer: the same linear operator as $D_1$, but expressed on the basis given by the columns of $A^{-1}$. So this matrix is diagonal if and only if every column of $A^{-1}$ is an eigenvector of (the linear operator on $K^n$ defined by multiplication by) $D_1$. But since $A$ is a permutation matrix, so is $A^{-1}$ (for the inverse permutation), and its columns are just the standard basis vectors of$~K^n$, in some order, and we just saw that each of these is an eigenvector of$~D_1$. QED.
