Given $A_{n\times n},B_{n\times n} \in \mathbb C$ then:
if $A$ is unitary and the characteristic polynomial $f_A(x)=f_B(x)$ then $B$ is also unitary.
if $A$ is normal and $f_A(x)=f_B(x)$ then $B$ is also normal.
if $A$ is unitary and $f_A(x)=f_B(x)$ and $m_A(x)=m_B(x)$ then $A$ is similar to $B$.
(1) if the matrix is unitary then the eigenvalues are $\pm 1$ but why would having eigenvalues that are $\pm 1$ mean that the matrix is unitary? So that's probably false.
(2) similar reasoning, why would the normal property carry over because of the same eigenvalues? Probably false.
(3) since $A$ is diagonalizable because its unitary, then it has a similar diagonal matrix, and since $B$ has the same diagonal matrix then it has to be similar to it? How does the minimal polynomial help here?
The characteristic polynomial of a linear map is defined without any additional structure beyond that of a finite dimensional vector space while the notion of unitarity or normality requires the structure of an inner product so you should be suspicious if someone tells you that the characteristic polynomial tells you something about unitarity or normality.
Indeed, if
$$ A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \,\,\, B = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} $$
then $A$ and $B$ have the same characteristic polynomial $(x - 1)^2$, $A$ is unitary and normal while $B$ isn't (for example, because it is not diagonalizable).
Regarding (3), the fact that $m_A(x) = m_B(x)$ implies that $m_B(x)$ splits into distinct linear factors and thus $B$ is diagonalizable. Without knowing that $B$ is diagonalizable, you wouldn't be able to deduce that $A$ and $B$ are similar - again, see the example above.