We consider $n\times n$ complex matrices. Let $i\_+(A), i\_-(A), i\_0(A)$ be the number of eigenvalues of $A$ with positive real part, negative real part and pure imaginary. It is well known if two Hermitian matrices $A$ and $B$ are $\*-$congruent, then
 $$(i\_+(A), i\_-(A), i\_0(A))=(i\_+(B), i\_-(B), i\_0(B)).\qquad{(1)}$$
If two general matrices $A$ and $B$ are $\*-$congruent, (1) may not hold (can you provide an example?).
Moreover, whether a matrix and its transpose are always $\*-$congruent?



A question on star-congruence.
An answer to the second question: Yes, a square complex matrix is always $\*$-congruent to its transpose, according to a more general result proved by Horn and Sergeichuk in "[Congruences of a square matrix and its transpose](http://arxiv.org/abs/0709.2489)". They prove the result for all fields with involution in characteristic other than 2.


*Added:*
Here's a counterexample for your first question:


$\left( \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right) \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right) \left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right) = \left( \begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix} \right)$.


