= Solution
For a partition $\nu$ with at most $m$ parts, let $D_\nu(V)$ be the <Schur module> constructed in Question 3, equivalently the image of a <Young symmetrizer> on $V^{\otimes|\nu|}$. For a weakly decreasing integer tuple $\lambda$, put $k=\max(0,-\lambda_m)$ and $\nu=\lambda+k(1,\ldots,1)$. Define the <rational Schur module> by
$$
\boxed{D_\lambda(V)=\det^{-k}\otimes D_\nu(V).}
$$
Here $\det^{-k}$ is the one-dimensional representation $g\mapsto\det(g)^{-k}$. This is rational, and it is irreducible under the permitted irreducibility assumption; for nonnegative $\lambda$ it is the original <polynomial> <Schur module>. Larger shifts give the same module, as will also follow from the <character> formula below.
Write $s_r(x)=\sum_{i=1}^m x_i^r$ and $p_\alpha(x)=\prod_{r=1}^n s_r(x)^{\alpha_r}$. For a <permutation> $\sigma$ of the given cycle type, the <trace of a permuted tensor power> is
$$
\operatorname{tr}_T(\sigma\xi^{\otimes n})
=\prod_{r=1}^n\operatorname{tr}(\xi^r)^{\alpha_r}=p_\alpha(x).
$$
In a tensor basis, the trace contracts the matrix entries of $\xi$ around each cycle of $\sigma$; a cycle of length $r$ contributes $\operatorname{tr}(\xi^r)$. This proof applies to nondiagonalizable <endomorphisms> as well. The <eigenvalues> give $\operatorname{tr}(\xi^r)=s_r(x)$.
The <Schur–Weyl duality> decomposition, on which $\sigma$ and $\xi^{\otimes n}$ act on the two respective factors, gives the same trace as
$$
\boxed{p_\alpha(x)=\sum_{\substack{\lambda\vdash n\\\ell(\lambda)\leq m}}
\chi^\lambda(\alpha)\phi_\lambda(\xi).}
$$
For a partition, the <character> extends polynomially to all <endomorphisms> because the tensor-power action does. This extension is not asserted for determinant-twisted modules at singular matrices.
We now derive the <alternant character formula for the general linear group>. Put $\delta=(m-1,m-2,\ldots,0)$ and
$$
a_\beta(x)=\det(x_i^{\beta_j})_{1\leq i,j\leq m},\qquad
a_\delta(x)=\prod_{i<j}(x_i-x_j).
$$
Use the permitted symmetric-group <character> result, the <Frobenius alternant character formula>:
$$
\chi^\lambda(\alpha)=[x^{\lambda+\delta}]\bigl(a_\delta(x)p_\alpha(x)\bigr).
$$
It concerns <characters> of $S_n$, rather than assuming the <character> formula we seek for the general linear group. Substitute the proved trace identity and set
$$
c_{\lambda\mu}=[x^{\lambda+\delta}]\bigl(a_\delta\phi_\mu\bigr).
$$
Then $\chi^\lambda(\alpha)=\sum_\mu c_{\lambda\mu}\chi^\mu(\alpha)$ for every <conjugacy class>. Independence of the irreducible symmetric-group <characters> forces $c_{\lambda\mu}=\delta_{\lambda\mu}$.
Each <character> $\phi_\mu$ on diagonal matrices is a symmetric homogeneous <polynomial> of degree $n$, since conjugation by <permutation> matrices permutes its arguments. Thus $a_\delta\phi_\mu$ is alternating of degree $n+\binom m2$. Every <alternating polynomial> of this degree has a unique expansion in alternants $a_{\lambda+\delta}$: a <monomial> with repeated exponents has zero coefficient, while each strictly decreasing nonnegative exponent vector is uniquely $\lambda+\delta$ for a partition $\lambda$ of $n$ with at most $m$ parts. Its coefficient at $x^{\lambda+\delta}$ is the coefficient of that alternant. The identities for $c_{\lambda\mu}$ therefore give $a_\delta\phi_\mu=a_{\mu+\delta}$. We have derived the <Weyl character formula>
$$
\boxed{\phi_\lambda(\operatorname{diag}(x_1,\ldots,x_m))
=\frac{\det(x_i^{\lambda_j+m-j})}{\det(x_i^{m-j})}.}
$$
For partitions the quotient is the <Schur polynomial>, with removable apparent singularities when <eigenvalues> coincide. For arbitrary dominant integer tuples, multiply the formula for $\nu$ by $(x_1\cdots x_m)^{-k}$; this shifts every numerator exponent by $-k$ and yields the same boxed formula. The variables must then be nonzero. The expression also shows independence of the shift used to define the <determinant twist>. Equality of <characters> identifies these <irreducible modules>: the group-algebra image on a <direct sum> of two irreducibles is finite-dimensional and semisimple, and its span of group operators detects the traces on every simple block.
Finally, the <character> of a <dual representation> evaluates the original <character> at $x_i^{-1}$. Reverse the numerator's columns after making this substitution. The exponents become $-\lambda_{m+1-j}+1-j$. Factoring $(x_1\cdots x_m)^{-(m-1)}$ converts these into $\mu_j+m-j$ for $\mu_j=-\lambda_{m+1-j}$. The denominator undergoes the identical column reversal and factor, so both signs and factors cancel. Hence $\phi_\lambda(x^{-1})=\phi_\mu(x)$, and
$$
\boxed{D_\lambda(V)^*\cong D_{(-\lambda_m,-\lambda_{m-1},\ldots,-\lambda_1)}(V).}
$$
This tuple is again weakly decreasing, so it is precisely the required dominant label.
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