Consider the $4\times4$ matrices $A=\bigg[\begin{matrix} E&O_2\\O_2&F \end{matrix}\bigg]$, where $E,F$ are any nilpotent $2\times2$ matrices, and $B_b=\left[\begin{matrix}2b&-1&0&-1\\0&b&0&0\\0&-1&0&-1\\0&1&0&b \end{matrix}\right]$ with $b\in\Bbb R$. For what values of $b$ are $A$ and $B_b$ similar?
I first checked a necessary condition: the characteristic polynomials must be equal.
By the nilpotence of $E$ and $F$, $$\det(A-\lambda I)=\det(E-\lambda I) \cdot \det(F-\lambda I)=\lambda^2\cdot\lambda^2=\lambda^4$$and $$\det(B_b-\lambda I)=\lambda(\lambda-2b)(\lambda-b)^2 $$so they are equal if and only if $b=0$.
Now, minimal polynomials should also be the same, and I noticed $A^2=\bigg[\begin{matrix} E^2&O_2\\O_2&F^2 \end{matrix}\bigg]=0,$ so the minimal polynomial of $A$ is $x^2$. But then $(B_0)^2$ is not the null matrix, the minimal polynomial of $B_0$ appears to be $x^3$, which would then mean $A$ and $B_b$ are not similar for any $b\in\Bbb R$.
But it seems strange, did I make a mistake?
You could replace $B_b$ by a similar triangular matrix to see somewhat better what is going on. Here you could just permute the standard basis vectors into the order $[e_3,e_1,e_4,e_2]$ to give a matrix $$ B'_b=\pmatrix{0&0&-1&-1\\0&2b&-1&-1\\0&0&b&1\\0&0&0&b} $$ similar to $B_b$, and which is nilpotent only for $b=0$. Since the final standard basis vector now survives two multiplications by $B'_0$ before becoming zero: $$(0,0,0,1)\mapsto(-1,-1,1,0)\mapsto(-1,-1,0,0)\mapsto(0,0,0,0),$$ it is clear that $B'_0$, unlike $A$, is not annihilated by the polynomial $X^2$, so $A$ and $B'_0$ are not similar. Your conclusion that $A$ is not similar to any $B_b$ is correct.