A question about the dual representation. Consider a representation $\rho : G \to \operatorname{GL}(V)$. Then we know that a dual representation $$\rho^* : G \to \operatorname{GL}(V^*)$$ is given by $\rho^*(g) = \rho(g^{-1})^t$. Now, consider some $\alpha\in V^*$. Is it clear that
$$\alpha(\rho(g^{-1})v)=(\rho(g^{-1}))^t\alpha(v)?$$
Does it follow from the fact that $T^*f=f\circ T$ for all $f\in V^*$ and $T \in \operatorname{Hom}(V,V)$?
 A: The dual representation is, as its name suggests, naturally defined on the dual $V^{\ast}$, not on $V$; that is, it's $\rho : G \to GL(V^{\ast})$. The fact that you mix taking the abstract dual with performing the concrete transpose operation makes your question slightly ill-posed, and unmixing and picking one or the other will make things clear.
Let's get rid of the transpose and work with abstract duals. The dual of a linear map $T : V \to W$ between vector spaces is the linear map $T^{\ast} : W^{\ast} \to V^{\ast}$ defined on linear functionals $f : W \to k$ by
$$T^{\ast} : W^{\ast} \ni f \mapsto (v \mapsto f(Tv)) \in V^{\ast}.$$
An equivalent way to phrase this definition, which among other things makes the relationship to taking transposes transparent, is that $T^{\ast}$ satisfies
$$\langle T^{\ast}(f), v \rangle_V = \langle f, T(v) \rangle_W$$
where $\langle -, - \rangle_V$ is suggestive notation for the dual pairing $V^{\ast} \otimes V \to k$.
If $\rho : G \to GL(V)$ is a representation of a group $G$ on a f.d. vector space $V$, we get a dual representation $\rho^{\ast} : G \to GL(V^{\ast})$ via
$$\rho^{\ast}(g) = \rho(g^{-1})^{\ast}.$$
This means by definition that
$$\langle \rho^{\ast}(g)(f), v \rangle_V = \langle f, \rho(g^{-1})(v) \rangle_V$$
which, if I've understood correctly, is what you're asking for but with the LHS and RHS switched.
