So, where $\lambda$ is the eigenvalue, and $A$ is the transform matrix:

$\displaystyle AX=\lambda IX$

$\displaystyle (A-\lambda I)X = 0$

Since X can only be 0, we know that $(A-\lambda I)$ must be rank 4 and therefore $(A-\lambda I)$ is row equivalent to $I$. Therefore you need $A$ to be some matrix that stays rank 4 no matter what you add/subtract from the diagonals.

$A= \begin{bmatrix}

0 & 1 & 0 & 0 \\

0 & 0 & 2 & 0 \\

0 & 0 & 0 & 3 \\

4 & 0 & 0 & 0

\end{bmatrix} $

Works since no matter what lambda is the matrix won't have any empty rows or columns. In this case $T = AX$

EDIT: Made a mistake in the original, if there are only $1$'s vectors with all equal elements will be eigenvectors. Apologies