Conjugating a non-identity Pauli operator $P$ by Clifford operators yields all non-identity Pauli operators, including $P$, with equal frequency.
Proof Let $G_n$ and $C_n$ denote the $n$-qubit Pauli and Clifford groups, respectively. For $P, Q\in G_n$, let $C_{P\to Q}$ denote the set of Clifford operators that map $P$ to $Q$ under conjugation, i.e.
$$
C_{P\to Q}=\{U\in C_n\,|\,UPU^\dagger=Q\}.
$$
For any $P\in G_n$, $C_{P\to P}$ is a subgroup of $C_n$. Indeed, for any $U\in C_{P\to P}$ its inverse$^1$ $U^\dagger\in C_{P\to P}$ and for any $U, V\in C_{P\to P}$ their product $UV\in C_{P\to P}$.
Let $P,Q\in G_n$ and suppose that neither $P$ nor $Q$ is the identity$^2$. Then there exists$^3$ $V\in C_{P\to Q}$. It is easy to see that
$$
C_{P\to Q} = VC_{P\to P}
$$
where $VC_{P\to P}$ is the left coset
$$
VC_{P\to P} := \{VU\,|\,U\in C_{P\to P}\}
$$
of the subgroup $C_{P\to P}$ in $C_n$. By group theory, all cosets have the same size, so
$$
|C_{P\to Q}| = |C_{P\to P}|
$$
for all $P,Q\in G_n$. Conclusion follows from the observation that $|C_{P\to Q}|$ is precisely the number of Cliffords that map $P$ to $Q$ under conjugation.$\square$
$^1$Strictly speaking, we do not need to check for closure under inverses. Explanation why is an exercise for the reader.
$^2$ The assumption is important, because $C_{I\to I}=C_n$ and $C_{I\to R} = C_{R\to I} =\emptyset$ for all non-identity $R\in G_n$.
$^3$ Proof that such $V$ exists is another exercise for the reader.
