Is every Clifford gate conjugate to a diagonal Clifford gate?

Question

Let $ C $ be a Clifford gate. Let $ D $ be the diagonalization of $ C $. In other words $ D $ is a diagonal gate and $$ C=VDV^{-1} $$ for some $ V $. Is $ D $ also a Clifford gate?

Update: Filling in the details of my comment below.

Claim: There exists a Clifford gate whose diagonalization is not in any level of the Clifford hierarchy.

Proof: Let $ H $ be the hadarmard and $ P=diag(1,i) $ the phase gate. Then $ M:=(H*P)^8 $ is a Clifford gate of order 3. Recall that (up to global phase) a diagonal matrix in the $ k $ level of the Clifford hierarchy has all its entries $ 2^k $ roots of unity. See https://arxiv.org/abs/1608.06596 . So every diagonal matrix in the Clifford hierarchy is a global phase times a matrix with order $ 2^k $. Thus a diagonal matrix which is in the Clifford hierarchy but has order relatively prime to $ 2^k $ must be a global phase times the identity (a unitary scalar matrix).

Suppose for the sake of contradiction that the diagonalization of $ M $ is in the Clifford hierarchy. Since $ M $ has order 3 and is not a scalar matrix then $ D $ must also have order 3 and also not be a scalar matrix. So this non scalar diagonal matrix $ D $ of order prime to $ 2^k $ contradicts the fact above.

Answer 1

No, $D$ isn't guaranteed to be Clifford. That would require all entries on the diagonal to differ by multiples of 90 degrees (all be 1, $i$, $-i$, or -1 up to global phase). But that would imply $D$ had period 1, 2, or 4 up to global phase. But there are Clifford gates that have periods other than 1, 2, or 4 up to global phase. For example, the 120 degree rotation around X+Y+Z is Clifford but has period 3.