Eigenvectors of a normal matrix According to the spectral theorem every normal matrix can be represented as the product of a unitary matrix $\mathbb{U}$ and a diagonal matrix $\mathbb{D}$
 $$\mathbb{A} = \mathbb{U}^H\mathbb{D}\mathbb{U}$$
meaning that every normal matrix is diagonalizable.
Does it necessarily mean that the unitary matrix has to be composed from the eigenvectors of $\mathbb{A}$ ? 
I presume that not, because then the eigenvectors of every normal matrix would form an orthonormal set (rows and columns of a unitary matrix are orthonormal in $\mathbb{C}^n$). So am I right that only the set of eigenvectors of a hermitian (or symmetric while in $\mathbb{R}^n$) matrix is orthonormal?
An $n\times n$ matrix $A$ over the field $\mathbf{F}$ is diagonalizable if and only there is a basis of $\mathbf{F}^n$ of eigenvectors of $A$. This occurs if and only if there exists an invertible $n\times n$ matrix $Q$ such that $Q^{-1}AQ$ is a diagonal matrix; the column of $Q$ form the basis made up of eigenvectors, and conversely, if you take a basis made up of eigenvectors and arrange them as columns of a matrix, then the matrix is invertible and conjugating $A$ by that matrix will yield a diagonal matrix.

An $n\times n$ matrix $A$ with coefficients in $\mathbb{R}$ or $\mathbb{C}$ is orthogonally diagonalizable if and only if there is an orthonormal basis of eigenvectors of $A$ for $\mathbb{R}^n$ (or $\mathbb{C}^n$, respectively).

An $n\times n$ matrix with coefficients in $\mathbb{C}$ is orthogonally diagonalizable over $\mathbb{C}$ if and only if it is normal; a square matrix is orthogonally diagonalizable over $\mathbb{R}$ if and only if it is Hermitian. "Unitary" is usually reserved for complex matrices, with "orthogonal" being the corresponding term for real matrices.
Here is what I think is correct:
Normal matrices are matrices that have orthogonal eigenvectors.
Hermitian matrices are normal matrices that have real eigenvalues.

So this answers your first question in positive: Yes, the unitary matrix in your decomposition has the same eigenvectors as your original matrix.