In the paper Computing Rational Forms of Integer Matrices by MARK GIESBRECHT† AND ARNE STORJOHANN
It says that, When A ∈ Qn×n has all integer entries, the Frobenius form F of A has all integer entries as well.
But how about if A ∈ Znxn is diagonalizable with eigenvalues not in Z ?
Then its diagonalized form is not equal to its rcf ?
The rational canonical form of a matrix is almost never a diagonal matrix. A typical rational canonical form is a companion matrix: $$\begin{pmatrix}0&0&0&\cdots&0&-a_0\\1&0&0&\cdots&0&-a_1\\ 0&1&0&\cdots&0&-a_2\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&0&-a_{n-2}\\ 0&0&0&\cdots&1&-a_{n-1}\end{pmatrix}$$ Even if the roots are all distinct and the matrix is diagonalizable, that's the form we'll choose to call the rational canonical form. When the RCF isn't of this form, it's because the minimal polynomial has lower degree than the characteristic polynomial - and then we build the RCF using companion matrix blocks. These companion matrices have integer entries when the associated polynomials have integer coefficients, which is certainly true for the minimal and characteristic polynomials of integer matrices.
In fact, the only way for the RCF to be diagonal is if the matrix is a multiple of the identity. In that case, there's no way for it to be similar to anything else.