Recall that similarity of matrices is an equivalence relation, that is, the relation is reflexive, symmetric and transitive.
Verify that $A=\left[\begin{array}{cc} 0 &-3\cr -3 &3 \end{array}\right]$ is similar to itself by finding a $T$ such that $A = T^{-1} A T$.
$T =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$
We know that $A$ and $B=\left[\begin{array}{cc} 0 &3\cr 3 &3 \end{array}\right]$ are similar since $A = P^{-1} B P$ where $P = \left[\begin{array}{cc} -1 &1\cr -1 &2 \end{array}\right]$. Verify that $B\sim A$ by finding an $S$ such that $B = S^{-1} A S$.
$S =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$
We also know that $B$ and $C=\left[\begin{array}{cc} -6 &15\cr -3 &9 \end{array}\right]$ are similar since $B = Q^{-1} C Q$ where $Q = \left[\begin{array}{cc} -3 &1\cr -1 &0 \end{array}\right]$. Verify that $A\sim C$ by finding an $R$ such that $A = R^{-1} C R$.
$R =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$

You can earn partial credit on this problem.