Conjugating matrix for diagonal form to companion matrix The question could be very elementary, but I was unable to think on it, and unable to get link through any appropriate sections in the books.
For simplicity, let $A$ be a $3\times 3$ diagonal matrix with distinct diagonal entries $a,b,c$, over a field say $\mathbb{R}$.
Let $B$ be the companion matrix of the polynomial $(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$
$$A=\begin{bmatrix} a & & \\ & b & \\ & & c\end{bmatrix} \,\,\,\,\,\,\,\,B=\begin{bmatrix} 0 & 0 & abc\\ 1 & 0 & -(ab+bc+ac) \\ 0 & 1 & a+b+c\end{bmatrix} $$
It can be seen that the Jordan form of matrix $B$ is nothing but $A$.

Is it easy to find (describe) a matrix $P$ such that $PBP^{-1}=A$ (or $PAP^{-1}=B$)?

I am considering this problem actually for general case (i.e. $n\times n$ case, over arbitrary field), but I am not getting proper reference for explicit form of a matrix $P$. For simplicity, I posed here 3 by 3 case; one may give hint also, instead of full answer. Or one may write sequence of statements, without proofs; I will try to prove them.
 A: 
Is it easy to find (describe) a matrix $P$ such that $PBP^{-1}=A$ (or $PAP^{-1}=B$)?

Yes: Let $P$ be the Vandermonde Matrix.  In particular
$P := \begin{bmatrix}
1 & a & a^2 \\ 
1 & b & b^2 \\ 
1 & c & c^2 \\ 
\end{bmatrix}$
$PBP^{-1} = A$
It's easy to check that the moment curve associated with a given eigenvalue (i.e. a row of $P$) is a left eigenvector of the Companion matrix and the result follows.
A: The columns of $P$ can be obtained from the eigenprojections of $B$. More specifically, if $A=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$ has $n$ distinct eigenvalues and $B$ is similar to $A$, then
$$
\Pi_i=\prod_{j\ne i}\frac{B-\lambda_j I}{\lambda_i-\lambda_j}
$$
is the projection operator onto the eigenspace of the eigenvalue $\lambda_i$. It fact, if for each $j$ the vector $v_j$ is an eigenvector for $\lambda_j$, one may verify directly that $\Pi_iv_j=\delta_{ij}v_j$. It follows that if $p_i$ is any nonzero multiple of any nonzero column of $\Pi_i$, then $B=PAP^{-1}$ where $P=\pmatrix{p_1&\cdots&p_n}$. E.g. when
$$
A=\pmatrix{1\\ &2\\ &&3},
\ B=\pmatrix{0&0&6\\ 1&0&-11\\ 0&1&6},
$$
we have
\begin{aligned}
(B-2I)(B-3I)
&=\pmatrix{6&6&6\\ -5&-5&-5\\ 1&1&1},\\
(B-I)(B-3I)
&=\pmatrix{3&6&12\\ -4&-8&-16\\ 1&2&4},\\
(B-I)(B-2I)
&=\pmatrix{2&6&18\\ -3&-9&-27\\ 1&3&9}.
\end{aligned}
Therefore we have $B=PAP^{-1}$ where
$$
P=\pmatrix{6&-3&2\\ -5&4&-3\\ 1&-1&1}.
$$
