Consider the following density matrix over $n$ qubits, with $C$ being a single qubit operator: $$ \rho_{n} = \int_{C \sim \text{Haar}} \big(C|0\rangle\langle0|C^\dagger\big)^{\otimes n} dC. $$
Let's say I measure the qubit with respect to orthogonal measurement operators $\{M_i : I \in [k] \}$. Then, for each $M_i$:
$$\text{Tr}\big(M_i \rho_n\big) = \int_{C \sim \text{Haar}}\text{Tr}\bigg(M_i \big(C|0\rangle\langle0|C^\dagger\big)^{\otimes n} \bigg) dC $$
$$= \int_{C \sim \text{Haar}} \text{Tr}\bigg(C^{\dagger \otimes n} M_i C^{\otimes n} \left(|0^{\otimes n}\rangle\langle 0^{\otimes n}|\right)\bigg) dC $$
$$= \text{Tr} \bigg(\overline{M}_i |0^{\otimes n}\rangle\langle 0^{\otimes n}| \bigg),$$
where we have
\begin{equation} \overline{M}_i = \int_{C \sim \text{Haar}} C^{\dagger \otimes n} M C^{\otimes n}. \end{equation}
Now note that by the left and right invariance of the Haar measure, for any one qubit unitary $U$,
$$U^{\otimes n} \overline{M}_i = \overline{M}_i U^{\otimes n}.$$
Then, by the Schur Weyl duality, with each distinct $\pi$ being a distinct permutation operator acting on the $n$ registers and for some choices of $a_{\pi} \in \mathbb{C}$
$$\overline{M}_i = \sum_{\pi \in S_{n}} a_{\pi} ~\pi.$$
Note that for any choice of $\pi$,
$$\pi |0^{n}\rangle = |0^{n}\rangle. $$
Then,
$$\text{Tr}\big(M_i \rho_{n}\big) = \text{Tr} \bigg(\overline{M}_i |0^{\otimes n}\rangle\langle 0^{\otimes n}| \bigg).$$
What is the relation between $a_{\pi}$ and irreducible representations of symmetric subspaces and weak Schur sampling? For example, on page 42 of this link (https://arxiv.org/pdf/1310.2035.pdf), it is written that we can write an operator like $\overline{M_i}$ as
$$ \overline{M_i} = \sum_{\lambda} a_{\lambda} P_{\lambda},$$
where $\lambda$ is a partition of $n$. How do I see the relation between these two representations?
Additionally, can we exploit symmetry properties, like the fact that
$$ \pi \overline{M_i} = \overline{M_i},$$
for any $\pi$, to say anything more about $a_{\pi}$ and $\overline{M_i}$?