$A, B$ in Jordan normal form, $A \neq B$, is it possible that $A$ is similar to $B$

Question

Let $A, B \in M(n \times n, \mathbb{K})$ be some matrices in Jordan normal form with $A \neq B$ (even after reordering of the jordan blocks). Is it still possible that $A$ is similar to $B$?

If think, if they both were in Frobenius normal form it would be impossible that $A$ is still possible to $B$, but how about the Jordan normal form?

Answer 1

Let $\text{GL}(n,\Bbb C)$ act on itself by conjugation. Then the orbits of this action are the similarity classes. Inside of each orbit, there is a matrix in Jordan normal form, with $k$ Jordan blocks of sizes $i_1\geq i_2\geq \dots\geq i_k$. Call this matrix $J$. Assuming $i_1\ne i_k$, one could rearrange these blocks, (remaining in Jordan normal form) and have a matrix $J'\ne J$.

Answer 2

They cannot be similar (if we assume equality up to ordering blocks). One argument is that one can determine the number of Jordan blocks for eigenvalue $\lambda$ from the dimensions of the nullspaces of $(A-\lambda)^k$ for $k=0,1,\ldots$.

These dimensions however are obviously invariant under similarity. Thus it is a genuine canonical form.