Let $n$ be a positive integer and $\lambda$ be a partition of $n$, which we identify with its Young diagram. Let $S^{\lambda}$ be the Specht module associated to $\lambda$.
Here the Specht modules are constructed via tabloids. More precisely, giving a numbering $T:\lambda\to [n]$, that is, a bijective function from the set of boxes of $\lambda$ to the set $[n]=\{1,2,\dots,n\}$, the symmetric group $S_n$ acts on $T$ by the formula $$ (\sigma\cdot T)(b) = \sigma(T(b)), $$ for each box $b\in \lambda$.
Let $C(T)$ be the subgroup of $S_n$ that preserves each column of $T$, and let $R(T)$ the subgroup of $S_n$ that preserves each row of $T$. The orbit of $T$ under the action of $R(T)$ is called a tabloid of shape $\lambda$ and is denoted by $\{T\}$. Let $M^\lambda$ be the complex vector space with basis the set of tabloids of shape $\lambda$. The symmetric group acts on the set of tabloids via $$ \sigma\{T\}=\{ \sigma T\}, \quad \sigma\in S_n $$ and this gives $M^\lambda$ the structure of a $\mathbb{C}[S_n]$-module. Let $$ v_T = \sum_{\sigma\in C(T)}\operatorname{sgn}(\sigma)\{\sigma T\}, $$ it is easy to see that $\sigma V_T = v_{\sigma\cdot T}$, thus the set $S^\lambda$ spanned by the vectors $v_T$ as $T$ runs over the set of tabloids of shape $\lambda$ is a $\mathbb{C}[S_n]$-submodule of $M^\lambda$ called the Specht module indexed by $\lambda$.
(I apologize, I now this is standard, but I want every answer to be given according to this precise construction. I know there are other realizations of $S^\lambda$, for example as ideals of the group algebra $\mathbb{C}[S_n]$).
Now my question:
Is there a $\mathbb{C}$-linear isomorphism $f:S^\lambda\to S^\lambda$ such that $$ f(\sigma v)=\operatorname{sgn}(\sigma) v $$ for all $\sigma\in S_n$ and all $v\in S^\lambda$? If so, what is an explicit description of such isomorphism in terms of the generators $v_T$?
What I know: From the isomorphism between the ring of symmetric functions and the ring of representations of the symmetric groups, we know that $$ S^{\lambda^t}\otimes_{\mathbb{C}} S^{(1^n)} \cong S^{\lambda},\tag{1} $$ where $(1^n)$ is the partition with each part equal to $1$, and $\lambda^t$ denotes the transpose of $\lambda$. It is easy to see that $S^{(1^n)}$ is the alternating representation of $S_n$, that is, as $\mathbb{C}$-vector spaces, $S^{(1^n)}\cong \mathbb{C}$ and for each $z\in\mathbb{C}$ and $\sigma\in S_n$, $$ \sigma\cdot z = \operatorname{sgn}(\sigma)z. $$ We identify $S^{(1^n)}$ with $\mathbb{C}$ as vector spaces. There is an obvious isomorphism of $\mathbb{C}$-vector spaces $$ g:S^{\lambda^t}\otimes_\mathbb{C}\mathbb{C}\to S^{\lambda}, \quad v_T\otimes 1 \mapsto v_{T^t} $$ where $T^t$ denotes the numbering $T^t(i,j)=T(j,i)$ for the box $(i,j)$ in row $i$ and column $j$ of $\lambda$, and $T$ is a numbering of shape $\lambda^t$. Provided I made no mistakes, I proved that $$ g(\sigma\cdot(v_T\otimes 1)) = \operatorname{sgn}(\sigma) \sigma g(v_T\otimes 1), $$ for all $\sigma\in S_n$ and all numbering $T$. For me, this means that an explicit description of $f$ is equivalent to give an explicit formula for the isomorphism (1), but I know no such explicit isomorphism.