No, that is not true. The matrices
[tex]A= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}[/tex]
and
[tex]B= \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}[/tex]
have the same eigenvalues (1 with multiplicity two) but are not unitarily equivalent because they do not have the same eigenvectors. A has every vector as eigenvector while B has only multiples of <1, 0> as eigenvectors.

Two matrices are "unitarily equivalent" if and only if they have the same eigenvalues and the same corresponding eigenvectors.

Halls is definitely wrong to say that they have to have the same eigenvectors. You just have to have the same number of linearly independent eigenvectors for every eigenvalue. You have three distinct eigenvalues. That means you don't have to compute the eigenvectors. Why?