= Solution
For $\ell\in V^*$ define $(g\ell)(v)=\ell(g^{-1}v)$. The representing matrix is the transpose of the inverse matrix on $V$, so the dual <character> is
$$
\boxed{\beta(g)=\alpha(g^{-1})=\overline{\alpha(g)}.}
$$
The last equality follows by averaging a <Hermitian inner product> to make the finite-group <group representation> unitary. Complex finite-group representations are isomorphic exactly when their <characters> agree. Hence $V$ is self-dual exactly when its <character> is real-valued.
The invariant tensor identification $(V\otimes V)^G\simeq\operatorname{Hom}_G(V^*,V)$ and <Schur lemma> show that, for irreducible $V$, the trivial multiplicity is zero unless $V^*\simeq V$, and one in that case. Dually, a self-duality isomorphism gives a nonzero invariant <bilinear form> $B$ on $V$. Its radical is an invariant subspace, so irreducibility makes $B$ nondegenerate. The space of invariant <bilinear forms> is one dimensional. Since the transpose $B^T$ is another such form, $B^T=cB$; transposing again gives $c^2=1$. Therefore $B$ is symmetric or alternating: in the negative-transpose case, $B(v,v)=-B(v,v)$ forces $B(v,v)=0$. Both types cannot coexist, since proportional nonzero forms cannot have both transpose signs.
In odd dimension an alternating matrix has <determinant> zero: $\det B=\det(-B^T)=(-1)^n\det B=-\det B$. Thus only the symmetric case is possible. Over $\mathbb C$ a nondegenerate <symmetric bilinear form> has a <basis> with matrix $I$: diagonalize it by congruence and rescale each nonzero diagonal entry. In that <basis> invariance says $\rho(g)^T\rho(g)=I$. Consequently
$$
\boxed{\rho(G)\text{ is conjugate to a subgroup of }O(n,\mathbb C)\quad(n\text{ odd, irreducible self-dual}).}
$$
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