The dual or contragradient representation from a vector space $V$ on $V^*$ (the dual vector space of $V$) is defined as the linear operator
$$ (\pi^{-1})^T(g): V^*\rightarrow V^*, $$
where I implicitly assumed that $\pi(g)$ is a "representation" of a group $G$ on $V$.
On the other hand, in Physics one usually define the "Hermitian inner product" on $V$ as the map $V\times V\rightarrow \mathbb{C}$ such that
$$ \langle \alpha u,v\rangle = \overline{\alpha}\langle u,v\rangle, $$
with $u,v\in V$ and $\alpha$ a complex number (the overline is the conjugate operation).
Hence if $V$ has this inner product, one can define the adjoint operator of some operator $L$ via
$$ \langle Lu,v\rangle = \langle u, L^\dagger v\rangle. $$
Now here's my question/confusion: let $V$ be a linear vector space of finite dimension and $G$ the group of unitary transformations that preserve the above inner product. An element $g$ of $G$ has a certain representation (choosing an orthonormal basis) which is a unitary matrix $U_g$ that acts over $V$. Then $U_g$ acts over dual vectors as $U^\dagger_g$, which is the conjugate transpose matrix of $U_g$.
On the other hand, the contragradient representation says that $U_g$ will act over dual vectors as $\overline{U}_g$.
Of course here I am assuming that $\pi(g)=U_g$. What I can't see is why the contragradient representation acts differently in the dual space compared to the conjugate transposed induced by the inner product. Why the contragradient representation is useful compared to the more "natural" conjugate transpose in Physics?