A question about the adjoint of the Adams operations on representation rings Let $G$ be a finite group, and $R(G)$ its representation ring over $\mathbb{C}$. We have the Adams operations $\psi^k:R(G)\rightarrow R(G)$, given on the level of characters by: $$\chi_{\psi^k{V}}(g)=\chi_V(g^k).$$
Since the power sums can be expressed as polynomials in the homogenous symmetric functions $h_n$, these $\psi^kV$ correspond to virtual representations given by integer polynomials in the symmetric powers of $V$.
By using these same polynomials in the symmetric powers, we can define Adams operations on the level of Burnside rings $\psi^k:A(G)\rightarrow A(G)$, such that these operations intertwine the natural map $A(G)\rightarrow R(G)$ induced by taking the free (virtual) vector space on the (virtual) $G$ set.
Since $R(G)$ has a nondegenerate quadratic form, we can take the adjoint of the $\psi^k$ map, call it $\nu^k$. Since it is an adjoint, this map preserves characters/virtual representations. On characters, this map is given by:$$\chi_{\nu^k V}(g)=\sum_{h^k=g}\chi_V(h).$$
My question is then, does a natural lift of $\nu^k$ to the Burnside ring exist?
It is natural condition for such a $\nu^k$ to commute with induction from subgroups, so it suffices to define $\nu^k(\ast)$ for the trivial $G$ set $\ast$. So to show that such a Burnside ring version of this map isn't possible, it would suffice to show that $\nu^k(\mathbb{1})$ isn't in the image of the natural map $A(G)\rightarrow R(G)$, but I'm struggling to find a counterexample to this claim.
 A: If I understand the question correctly, then $\nu^{2}(1)$ isn't in the image of the natural map $A(G) \to R(G)$ when $G = Q_{8},$ the quaternion group of order $8$.
The number of square roots of the identity in $G$ is $2$, the number of square roots of the central involution $z$ in $G$ is $6$, and the number of square roots of each element of order $4$ is $0$.
Hence we have $\nu^{2}(1) = \lambda_{1} + \lambda_{2}+ \lambda_{3} + \lambda_{4} - \chi$, where the $\lambda_{i}$ are the linear characters of $G$ and $\chi$ is the unique irreducible character of $G$ of degree $2$.
Now I claim that this virtual character is not a difference of permutation characters. Indeed, it is not even a difference of characters afforded by $\mathbb{R}G$-modules.
For the irreducible character $\chi$ has real Schur index $2$, so occurs with even multiplicity in any character of $G$ afforded by an $\mathbb{R}G$-module.
Hence $\chi$ occurs with even multiplicity in any difference of permutation characters, so that $\nu^{2}(1)$ is not expressible as a difference of permutation characters.
More generally, if $G$ is any finite group of even order which has an irreducible character $\chi$ with Frobenius-Schur indicator $-1$, then $\nu^{2}(1)$ is not expressible as a difference of permutation characters of $G$.
An induction theorem of G. Segal may be relevant to trying to characterize exceptions for which $\nu^{2}(1)$ is not a virtual permutation character of $G$.
