Let $ P $ be a Pauli matrix. Pauli matrices are normal. So by the spectral theorem $ P $ can be written as $$ P=VDV^{-1} $$ for $ V $ unitary and $ D $ diagonal (in other words $ P $ is unitarily diagonalizable). Can we conclude that $ D $ must be in the Pauli group? Moreover, can we conclude that $ V $ must be in the Clifford group?
For certain examples of $ P $ and certain spectral decompositions this is obviously true. For example $$ X=HZH $$ here it is indeed the case that $ Z $ is a Pauli and $ H $ is in the Clifford group.
Summary:
The answer by DaftWullie shows that every Pauli matrix is diagonalizable by a Clifford gate $ V $. This implies (indeed is equivalent to) the fact that the diagonalization of a Pauli matrix is always another Pauli matrix.
Discussion with Bebotron shows that there is no uniqueness to this diagonalizing gate $ V $. A given Pauli matrix can be diagonalized by both Clifford and non Clifford gates. And multiple different Clifford gates can diagonalize the same Pauli. DaftWullie comments further on the extraordinary non-uniqueness of the diagonalizing gate $ V $.