Example of 2 matrices similar but not row equivalent

$$\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix}$$

$\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}$ and $\begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix}$ are not row equivalent though they are similar.

Consider all $n\times n$ diagonal matrices whose diagonal entries are $n$ distinct fixed numbers $d_1,d_2,\ldots, d_n$. (there are $n!$ of them). Assume these numbers to be non-zero. Then it is easy to see that they are all similar to each other. (Use permutation matrices as done in the answer of Siong Thye Goh).

But these matrices are not row equivalent: two row equivalent matrices will lead to systems of linear equations with identical solutions. Let RHS vector be $v=(1,2,\ldots, n)^T$.

The system $DX= v$ and $D'X=v$ will not not have same solution for $D,D'$ diagonal matrices satisfying the conditions above.