For what $a,b,c,d$ are $A$ and $B$ similar? 
Consider the following two matrices with complex entries.
$$A = \begin{pmatrix} 1&a&0\\ 0&2&b \\ 0&0&2 \end{pmatrix}, B=\begin{pmatrix} 2&d&c \\ 0&c&d \\ 0&0&2 \end{pmatrix}$$
When are they similar?

My progress: First we see that $c$ has to be equal to $1$ if $A$ is similar to $B$. Characteristic polynomial of $A$ and $B$ is $(1-\lambda)(2-\lambda)^2$.
Suppose that the minimal polynomial of $A$ and $B$ are $(1-\lambda)(2-\lambda)$, then after computation, we get
$$(I-A)(2I-A)=\begin{pmatrix}0&0&ab\\ 0&0&b \\ 0&0&0 \end{pmatrix} \text{ and } (I-B)(2I-B)=\begin{pmatrix}0&0& 1+d \\0&0&0 \\ 0&0&0 \end{pmatrix}$$
So we conclude that if $d=-1$, then we must have $b = 0$. But I cannot track down $a$. Also what happens when their minimal polynomial is not $(1-\lambda)(2-\lambda)$?

You have made a mistake. It should be $1+d^2$ instead of $1+d$.
Matrices are similar iff they have the same Jordan form. The eigenvalues are $\{1,2,2\}$. The eigenvalue $1$ has, of course, a single scalar Jordan block, so the Jordan block for $2$ must have the same size for both. Since $\operatorname{rank}(B-2I)=2$ (assuming the matrices are real), we get that $\dim\ker(B-2I)=3-2=1$, so the Jordan block with eigenvalue $2$ in $B$ has the size $2\times 2$. For the matrix $A$ to have the same size of the corresponding block it must be $b\ne 0$. Therefore, the matrices are similar iff $b\ne 0$.
Edit: for complex entries we have $\operatorname{rank}(B-2I)=2$ iff $d^2+1\ne 0$. Thus the matrices are similar iff $d^2+1$ and $b$ are both zeros (the matrices are diagonalizable) or both non-zeros.
