Before anything else: the eigenvalues of B are 1 (twice) and 4, so you should first check your computation (or your post, more probably, since the eigenvalues of A are 1,2,3).

That said, the determinant and trace aren't enough: they give you the sum and the product of the three eigenvalues, and you would need a third information to find them.

I don't think you can avoid computing the roots of the characteristic polynomial. To show that the matrices are similar, you don't need to find the eigenvectors because the eigenvalues are different and hence the matrices are diagonalizable. To find the matrix , you'll need them however.