Is $V$ isomorphic to $V^*$ as representations of a finite group G? If $\dim(V)<\infty$ and $(V,\rho)$ is a representation of G, then $(V^*,\rho^*)$ is a representation of $G$ where $$\rho^*_g(f)(v)=f(\rho_{g^{-1}}(v)).$$
I tried to show they are equivalent using a basis $B_v=\{v_1,...,v_n\}$ and considering the linear transformation to its dual that sends $v_i$ to $v_i^*$ and I got really stuck. I guess this is not the way but I'm not sure either. Thanks for help or advice.
 A: Think of $(V,\rho)$ as a group of matrices.
If $(V,\rho)$ and $(V^*,\rho^*)$ are equivalent,
then $\exists B\in\text{GL}_n:\forall g\in G:B^{-1}\rho_g B=(\rho_{g^{-1}})^\tau$. Here $\tau$ means "transpose".
To find an example where $(V,\rho)$ and $(V^*,\rho^*)$ are not equivalent,
you could for instance look for cases where $\text{trace}(\rho_g)\neq\text{trace}(\rho_{{g}^{-1}})$.
A: I prefer the pairing notation: $(g(f), v) = (f, g^{-1}v)$.
Now if we pick basis $e_1, \cdots, e_n$ of $V$ and dual basis $e_1^*, \cdots, e_n^*$ of $V^*$. Then the we have $$\rho^*(g)_{ij} = (g(e_j^*), e_i) = (e_j^*, g^{-1}e_i) = \rho(g^{-1})_{ji}$$
In other words, if we have $\rho: G\rightarrow GL_n(F)$, then $\rho^*$ is equivalent to $\rho^*(g) = (\rho(g)^{-1})^t$.
These two representations are certainly not isomorphic in general. For example, take a nontrival $1$-dimensional representation of $\mathbb Z/3\mathbb Z$ such that $\rho(\bar 1)=e^{\frac{2\pi}{3}i}$, then $\rho^*(\bar 1) = e^{-\frac{2\pi}{3}i}$ which is apprrently different (hence nonisomorphic to ) $\rho$.
If we assume $\text{char} F \nmid |G|$, then the equivalency can be tested by comparing the traces, and apparently we have $\text{Tr}(\rho^*(g))=\text{Tr}(\rho(g^{-1}))\not=\text{Tr}(\rho(g))$ in general.
