Jordan Normal Form and its conjugates

Question

Is it true that a matrix of the form $$ \begin{bmatrix} 0 & x_1 & \ldots & \ldots & 0 \\ 0 & 0 & x_2 & \ldots & 0 \\ \vdots & \vdots & \ldots & \vdots & \vdots \\ 0 & 0 & \ldots & \ldots & x_{n-1} \\ 0 & 0 & \ldots & \ldots & 0 \end{bmatrix} $$ with nonzero $x_1,x_2, \ldots, x_{n-1}$ conjugate to the usual Jordan cell which has all $1$'s above the diagonal?

Answer 1

The answer is yes. For an explicit similarity, note that the Jordan block is given by $J = DMD^{-1}$, where $$ D = \pmatrix{1\\&x_1\\&&x_1x_2\\&&&\ddots\\ &&&&x_1x_2\cdots x_{n-2}x_{n-1}}, \\ M = \pmatrix{0 & x_1 & \ldots & \ldots & 0 \\ 0 & 0 & x_2 & \ldots & 0 \\ \vdots & \vdots & \ldots & \vdots & \vdots \\ 0 & 0 & \ldots & \ldots & x_{n-1} \\ 0 & 0 & \ldots & \ldots & 0}. $$