Is it that only with normal matrices, the transition matrix to its [del: inherent] [ins: own] basis is unitary? Does this even make sense what I translated into english?
PS. I am probably gonna delete this question eventually
 A: Jose has already given the answer, but here's a quick proof of it.  First, note that any diagonal matrix D is normal, since its adjoint is also diagonal and diagonal matrices commute.  Now suppose you have some matrix A that is UDU* for U unitary and D diagonal, i.e. it is conjugate to a diagonal matrix by a unitary change of basis.  Then A*=(UDU*)*=UD*U*, so AA*=UDU*UD*U*=UDD*U*=UD*DU*=UD*U*UDU*=A*A, so A is normal.
Here's a more conceptual explanation of what's going on.  Normality is a property of a linear transformation on an inner product space.  If you pick an orthonormal basis, any transformation that is diagonal with respect to that basis is easily seen to be normal.  To say that A is conjugate to a diagonal matrix by a unitary matrix is exactly saying that A is diagonal with respect to some orthonormal basis, since a unitary matrix is just changing from the standard orthonormal basis to some other orthonormal basis.
A: I am not sure I understand the question, but if you are talking about linear algebra over the complex numbers, then it is true that normal matrices (those which commute with their hermitian adjoint) are precisely those which can be diagonalised by a unitary transformation.
(This is proven in Herstein's Topics in algebra: §10 of the 1964 edition.)
A: I think I understand the question in Serbian (my native language is from the same Slav family) and he is really asking about normal matrices and their diagonalization by a unitary transformation.
A: Other people have already answered the mathematical question. I'll point out a possible source of linguistic confusion: the word "eigen", in eigenvector and eigenbasis, is German for something like "own" or "self". I would guess that the poster may have been translating "eigen" into English rather than leaving it alone.
