The multipartition of the dual of a $C_m \wr S_n$ module I know that the irreducible modules of $C_m \wr S_n$ over $\mathbb{C}$ are parametrised by m-multipartitions. The parts of the multipartition are indexed by the elements of $C_m$.
My question now is: If $V$ is an irreducible module over $\mathbb{C}$ with multipartition $\underline{\lambda} = (\lambda_0, \lambda_1, ..., \lambda_{m-1})$, what is the multipartition of $V^*$?
I feel like it should be a permutation of $\underline{\lambda}$ corresponding to the map $i \mapsto m-i$, but I cannot find that statement anywhere. It should be fairly easy to prove but I have not managed to do so yet.
If the proof turns out to be too complicated, any source would suffice.
Thanks in advance!
 A: I assume that by a multipartition you mean here that $\lambda_i$ is a partition of $k_i$ and $\sum_i k_i = n$. In this case the correspondence you described is indeed the duality correspondence. I believe the easiest way to see this is via Clifford Theory: Consider the short exact sequence $$1\to C_m^n\to G\to S_n\to 1$$ where $G=C_m\wr S_n$. 
Let $\zeta$ be an $m$-th root of unity and let $g$ be a generator of $C_m$. Every irreducible representation of $C_m^n$ is of the form $$\rho_{t_1,\ldots,t_n}\big((g^{a_1},\ldots,g^{a_n})\big)=\zeta^{\sum_i t_ia_i}$$ 
where $t_i\in \{0,\ldots, m-1\}$. 
Every irreducible representation is $S_n$-conjugate to a unique representation in which $t_1\leq t_2\ldots \leq t_{n-1}$. Fix such a tuple. For $i\in \{0,\ldots, m-1\}$ write $k_i:= |\{j| t_j=i\}|$. The stabilizer of $\rho_{t_1,\ldots, t_n}$ in $S_n$ will then be isomorphic to $S_{k_0}\times\cdots\times S_{k_{m-1}}$. If $(\lambda_i,\ldots, \lambda_m)$ is a multipartition in the sense that $\lambda_i$ is a partition of $k_i$, then the corresponding tensor product of Specht modules $\mathbb{S}_{\lambda_0}\otimes\cdots\otimes \mathbb{S}_{\lambda_{m-1}}$ is an irreducible representation of $S_{k_0}\times\cdots\times S_{k_{m-1}}$. By letting $C_m^n$ act via the character $\rho_{t_1\ldots,t_n}$ on this vector space you get a representation $V$ of $$H:=C_m^n\ltimes (S_{k_1}\times\cdots\times S_{k_m}).$$ By taking the induced representation $\text{Ind}^G_H V$ you get a representation of $G$. Clifford Theory asserts that this is an irreducible representation of $G$, and that you get all the irreducible representations of $G$ this way.
Now for the duality: it holds that $$(\text{Ind}^G_H V)^* \cong \text{Ind}^G_H (V^*).$$
Since all the irreducible representations of $S_{k_0}\times\cdots\times S_{k_{m-1}}$ are self dual, you are left with the same representation of this group. However, $C_m^n$ acts now via the dual character of $\rho_{t_1,\ldots, t_n}$, which is $\rho_{m-t_1,\ldots, m-t_n}$. When you calculate the numbers $k'_i$ which correspond to this representation you will get now that $k'_i=k_{m-i}$. 
So the multipartition which corresponds to the dual representation is indeed $(\lambda_0,\lambda_{m-1},\ldots, \lambda_1)$
