How many eigenbases can a Hermitian matrix have? I learned in quantum mechanics that the eigenvectors of a Hermitian operator (matrix) can be made to form an orthogonal eigenbasis for the vector space. 
I also learned that 2 commuting Hermitian operators share a common eigenbasis. Does this mean that Hermitian operators can have more than one eigenbasis? In general, how many eigenbases can a Hermitian operator have? 
 A: arbitrarily many, just take the matrix corresponding to the scalar multiplicattion, this has every basis as an eigenbasis
A: If the eigenvalues of a diagonalizable operator $A$ all have multiplicity $1$ then the eigenbasis is unique up to multiplication of the eigenvalues by scalars. There is essentially just one basis of eigenvectors.
Any operator commuting with $A$ will have the same eigenvectors as $A$. To see why, suppose $v$ is an eigenvector for $A$. Then
$$
ABv = BAv = B\lambda v = \lambda Bv
$$
so $Bv$ is a $\lambda$-eigenvector of $A$. Since the space of those eigenvectors is one dimensional and spanned by $v$, $Bv$ must be a multiple of $v$, hence an eigenvector of $B$ for some eigenvalue.
I suspect that the multiplicity $1$ assumption is implicit in the quantum mechanical application you have in mind.
A: It is not true that two commuting Hermitian operators share a common eigenbasis, as you can see in this question, but if two operators share an eigenbasis, that is, they are simultaneously diagonalizable, they must commute.
A: The identity $I$ has a lot eigen bases. Any time an eigenspace has dimension $> 1$, there are different eigen bases. If the multiplicities of the eigen bases are all $1$, then your basis elements are unique up to unimodular constant multipliers.
