# Does conjugation by a Clifford send each non-identity Pauli to every other non-identity Pauli with equal frequency?

I see here in Olivia DeMatteo's notes, she states:

When we consider the action of the entire Clifford group on a single non-identity Pauli, it maps that Pauli to each of the $$d^2 − 1$$ other possible Paulis an equal number of times. Since we have $$|C_n|/|P_n|$$ possible Cliffords, we get mapped to each Pauli $$\frac{|Cn|/|Pn|}{d^2−1}$$ times.

This is also stated in this paper

[...] conjugation under the Clifford group maps each nonidentity Pauli element to every other nonidentity Pauli element with equal frequency.

I don't see a proof for this claim anywhere. Does anyone know of a proof for it? I think the proof for the cardinality of the Clifford group can be used to prove it, but I'm sure there is a simpler way to prove it.

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.