Isomorphism of faithful representations Let $G$ be a group and $f,g: G \rightarrow GL(V)$ be two faithful representations over some field $K$ with $f:x\mapsto f(x)$ and $g:x \mapsto f(x^{-1})$.
I would like to find out if $f$ and $g$ are isomorphic. So I need an ismomorphism $\alpha: V \rightarrow V$ with
$$\alpha \circ f(x) \circ \alpha^{-1} = f(x^{-1}). $$
What could such an isomorphism be? Does it depend on $K$?
Thank you for hints.

Later edit. As it is not possible to solve this in general: Is there an answer to the example $G=C_p$ cyclic group, $K$ some field of characteristic zero, $f,g$ irreducible representations of degree $> 1$?
 A: It depends. There are two separate issues here. Firstly, if $f$ is a faithful representation such that $g(x)=f(x^{-1})=f(x)^{-1}$ is also a representation, then the group $G$ is necessarily abelian:
$$f(x)^{-1}f(y)^{-1}=g(x)g(y)=g(xy)=f(xy)^{-1}=(f(x) f(y))^{-1}$$ implies that $f(x)$ and $f(y)$ commute and hence that $G$ is abelian. 
Secondly, some representations (for instance, the inclusion of $\{\pm 1 \}$ in the multiplicative group of your field) have the property that $f$ and $g$ are isomorphic and others do not---for instance, the defining representation of the group of third roots of unity (working over the complex numbers) is not isomorphic to its dual $g$.
In response to the edited question, calculate the character. When working with subfields of the complex numbers, the representation will be self-dual precisely when the character is real-valued.
A: If $f$ is faithful and $g$ is a homomorphism of groups, then $G$ is abelian. It follows that $V$ is a direct sum of representations of degree $1$, and it is easy to see that it is enough to suppose that $V$ is of degree $1$, for the general case follows from this.
Can you do that case?
