Show that a representation of a finite group is isomorphic to its dual if its character takes only real values This appeared as a part of showing that a representation of a finite group is isomorphic to its dual if and only if its character takes only real values. The "only if" part was easy to show. For the "if" part I tried to find an isomorphism of representations $\sigma: V \rightarrow V'$, where $V$ is the space of representation and $V'$ it's dual.
When I fixed a base $(e_1,...,e_n)$ that is orthonormal with respect to the product that makes the representation orthogonal and defined $\sigma$ as bringing each element of this base to it's dual, I found:
$$
(\sigma \circ g (e_i))(e_j) = [g]_{i,j}
$$ 
and 
$$
(g' \circ \sigma(e_i))(e_j) = \overline{[g]_{i,j}}
$$ 
where $[g]_{i,j}$ is the $(i,j)$ element of the representation of $g$ as a matrix under the basis $(e_1,...,e_n)$ and these two do not seem to be equal.
I tried to define $\sigma$ in other ways but none of them worked. Any help with this would be appreciated. 
 A: The statement follows easily form the following two observations:
1) Two representations are isomorphic if and only if they have the same character function.
2) The character function of the dual representation is the complex conjugate of the character function of the original representation.
1 is a fundamental result of character theory, which you should have seen somewhere in your course. If not, let me know what you do know about characters (orthogonality? Mashke's theorem? ...) and I'll see if I can explain it.
For 2, consider an element $x$ of your group $G$, let $g: G \rightarrow GL_{n}(\mathbb{C})$ be a representation of $G$. As $x$ has finite order, so does $g(x)$. As a result, there exists a basis vor $\mathbb{C}^{n}$ for which $g(x)$ is a diagonal matrix with norm 1 elements on the diagonal. The dual representation $g^{*}$ has matrix $g(x)^{-T}$ with respect to the same basis, which yields $\overline{g(x)}$ in the case of a diagonal matrix with only norm 1 elements. It follows that the character of $g^{*}$, evaluated in $x$, is the complex conjugate of the character of $g$ evaluated in $x$.
Hope this helped. Let me know if you need more clarification.
