Let $\mathbb{k}$ be a commutative ring, $G$ be a group, and $\chi: G \to \mathbb{k}^\times$ be a character of $G$ (a group homomorphism from $G$ to the group of units of $\mathbb{k}$). Then, given any $\mathbb{k}[G]$-module $V$, we may define the $\chi$-isotypic subspace $$ V^\chi = \{ v \in V \mid gv = \chi(g)v \text{ for all } g \in G\}$$ which is the maximal semisimple submodule made up of representations isomorphic to $\chi$. Dually, we may define the $\chi$-isotypic quotient $$ V_\chi = V / I_\chi, \quad \text{where } I_\chi = \{gv - \chi(g)v \mid v \in V, g \in G\}$$ which is the maximal semisimple quotient made up of representations isomorphic to $\chi$. For example, when $\chi$ is the trivial character, we recover the subspace of invariants and the quotient of coinvariants. There is a natural composition of maps $$ f_\chi: V^\chi \hookrightarrow V \twoheadrightarrow V_\chi$$ and we may ask: under what conditions is $f_\chi$ an isomorphism? Less specifically, under what conditions are there "natural" isomorphisms between $V^\chi$ and $V_\chi$?
For instance, if $V^\chi$ is a direct summand of $V$ (as $\mathbb{k}[G]$ modules), then it is not difficult to see that $I_\chi$ is its complement, and hence the composition $$ f_\chi: V^\chi \hookrightarrow V^\chi \oplus I_\chi \twoheadrightarrow V/I_\chi$$ is an isomorphism. So a sufficient condition for $f_\chi$ to be an isomorphism is semisimplicity of $V$. However, there are some natural cases where asking for $V$ to be semisimple is asking too much.
For example, let $V = U^{\otimes n}$ for some finite-rank free module $U$, and let the symmetric group $G = S_n$ act on $V$ by permuting tensor factors. Pick the sign character $\chi: S_n \to \mathbb{k}^\times$, $\chi(\sigma) = \mathrm{sgn}(\sigma)$. Then $$ V^{\mathrm{sgn}} = \{u_1 \otimes \cdots \otimes u_n \in V \mid u_{\sigma(1)} \otimes \cdots \otimes u_{\sigma(n)} = \mathrm{sgn}(\sigma) u_1 \otimes \cdots \otimes u_n \text{ for all } \sigma \in S_n\}$$ is the subspace of antisymmetric tensors, and $$ I_{\mathrm{sgn}} = \{u_{\sigma(1)} \otimes \cdots \otimes u_{\sigma(n)} - \mathrm{sgn}(\sigma) u_1 \otimes \cdots \otimes u_n \mid u_i \in U, \sigma \in S_n\}.$$ Provided that $2$ is invertible in $\mathbb{k}$, we have $$ I_{\mathrm{sgn}} = I_{\mathrm{alt}} = \{u_1 \otimes \cdots \otimes u_n \mid u_i \in U, u_i = u_j \text{ for some } i \neq j\}$$ and hence $$ V_{\mathrm{sgn}} = U^{\otimes n} / I_{\mathrm{sgn}} = U^{\otimes n} / I_{\mathrm{alt}} = \wedge^n(U)$$ The natural map $f_{\mathrm{sgn}}: V^{\mathrm{sgn}} \to \wedge^n(U)$ is only an isomorphism when $n!$ is invertible in $\mathbb{k}$. However, there is always an isomorphism $\wedge^n(U) \to V^{\mathrm{sgn}}$ given by $$ u_1 \wedge \cdots \wedge u_n \mapsto \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) u_{\sigma(1)} \otimes \cdots \otimes u_{\sigma(n)}$$ (We can check this is an isomorphism of free $\mathbb{k}$-modules by using facts about bases of $\wedge^n(U)$ and $U^{\otimes n}$). Is the reason this last map works an accident, or is there something more general going on?