Let
$$ B = \begin{bmatrix} a_{n-1} & a_{n-2} & \cdots & a_1 & 1 \\ a_{n-2} & a_{n-3} & \cdots & 1 & 0 \\ \vdots & \ \vdots & & \vdots & \vdots \\ a_{1} & 1 & \cdots & 0 & 0 \\ 1 & 0 & \cdots & 0 & 0 \end{bmatrix} \qquad \mathrm{and} \qquad 
F = \begin{bmatrix} 0 & 0 & \cdots & 0& -a_{n}\\ 1 & 0 & \cdots & 0 & -a_{n-1}\\ \vdots & \ \vdots & & \vdots & \vdots \\ 0 &  0 & \cdots & 0 & -a_{2} \\ 0 & 0 & \cdots & 1 & -a_1 \end{bmatrix}$$
We have $B^{-1}FB=F^T$. Then a companion matrix $F$
 and its transpose $F^T$
 are similar.