Invertibility

If $X^{-1}MX=I_n$ then $M=I_n$ (multiply with $X$ from the left and with $X^{-1}$ from the right). Thus, any invertible $M\neq I_n$ is a counterexample.

Let $M = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ . Since $\text{tr }M= 0$ , by the invariance of trace under conjugation, there is no invertible matrix $X$ such that $X^{-1} M X = I_n$ . But $M$ is clearly invertible.

Any triangular square matrix such that two or more element on the diagonal are the same is a counterexample. Such matrices are invertible. But, the eigenvector matrices of them are not invertible, which makes the original matrices not diagonalizable.