$$A=\begin{bmatrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \hspace{1cm} B=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix} \hspace{1cm} C=\begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
I would argue as follows:
Since all matrices are triagular, each has the same set of eigenvalues. As $B$ is diagonal, we can say that both $A$ and $C$ are diagonalizable, with their diagonal matrices being $B$. It follows that $A$ and $C$ are similar to $B$. By transitivity, they are also similar to each other. Hence all matrices are similar.
However, I was told that the order of the eigenvalues (it differs among the matrices) might invalidate this reasoning, but I don't see how that could be true, given that we may asign the order of the eigenvalues arbitrarily when diagonalizing a matrix.
On second though, I could just use the theorem that says that similar matrices have the same eigenvalues, but would that line of reasoning above still be valid?
HINT
Matrices $A$ and $B$ are similar indeed $B$ is the diagonal form of $A$.
Matrix $C$ is a Jordan canonical form (i.e. eigenvalues $2$ has geometric multiplicity $=1$).