Unit operator commutes with the parity operator i.e., $[\mathbb{1},\mathbb{P}]=0$. Any function $f(x)$ is an eigenfunction of the unit operator $\mathbb{1}$ while only even or odd functions are eigenfunctions of parity operator $\mathbb{P}$. I define unit operator as $\mathbb{1}f(x)=f(x)$. A parity operator is defined as $\mathbb{P}^2=\mathbb{1}$ or $\mathbb{P}f(x)=f(-x)$. Let the vector space be $L^2(\mathbb{R})$. Both $\mathbb{P}$ and $\mathbb{1}$ are hermitian in this space.
EDIT: In this space, do both operators have an equal number of linearly independent eigenfunctions? How do we prove that?
You do not need to invoke a theorem in order to prove that if an operator commutes with the identity operator, then they have common eigenvectors. That's trivial, since any (non-zero) vector is an eigenvector of the identity operator and any operator commutes with the identity operator.
So, the answer to your question is “yes”, but it's quite trivial. Indeed, there are infinitely many eigenvectors of the parity operator and each of them is an eigenvector of the identity operator. So, both operators have infinitely many eigenvectors.