Let $F=\mathbb{R}$ or $\mathbb{C}$.
I am wondering whether a set of invariants (characteristic polynomial, a set of eigen-values, the set of multiplicity of each eigen-values) exactly characterizes the similar relation of matrix.
More precisely, let $A,B\in M_{n \times n}(F)$ be the two matrices whose characteristic polynomials and total set of eigen-values are same. Furthermore, if $\lambda\in F$ is any common eigen-value of them, then we suppose that the algebraic multiplicity of $\lambda$ of $A$ and $B$ are same and ditto for their geometric multiplicities.
In this situation, can we say that $A$ and $B$ are similar? That is, is there an invertible matrix $U\in M_{n \times n}(F)$ such that $U^{-1}AU=B$?
I think it may not true. But I have some trouble in finding counter-example.
Any comments on this will be highly appreciated.
No, you can not. If you take all the invariants you gave, you have to make your matrices a little bigger, but there are still counterexamples (I think $8 \times 8$ should work, but I'm not sure). An invariant is the Jordan normal form. There is also a generalized Jordan normal form which will work over the reals. But taking characteristic polynomial, minimal polynomial, eigenvalues, multiplicities, etc. will not work if the matrices get to big. In fact, you can generate counterexamples by using Jordan normal forms, you just have to know how to read off all these invariants from the normal form.