Do characters distinguish real representations of a finite group?

Question

Let $G$ be a finite group. Consider two representations $\rho_1, \rho_2: G\to GL_{n}(\mathbb R)$. Suppose that these two representations have the same characters (i.e for any $g\in G$, ${\rm tr}\rho_1(g)={\rm tr}\rho_2(g)$). Is it true that these representations are isomorphic? If not, what is a simple counterexample?

I know that this statement holds for complex representations.

Answer 1

See the following answer of Derek Holt to a related question: link. A reference for what you need is Theorem 29.7 in Curtis and Reiner.

Let $G$ be a finite group and let $K \subseteq L$ be a field extension. Consider two representations $\rho_1: G \rightarrow \operatorname{GL}_n(K)$ and $\rho_2: G \rightarrow \operatorname{GL}_n(K)$ over $K$. If $\rho_1$ and $\rho_2$ are equivalent when regarded as representations $\rho_i: G \rightarrow \operatorname{GL}_n(L)$ over $L$, then $\rho_1$ and $\rho_2$ are equivalent.

In your question the two representations have the same character, hence they are equivalent over $\mathbb{C}$, and thus by the result mentioned they are equivalent over $\mathbb{R}$.

Answer 2

In characteristic zero, two representations are isomorphic over $K$ if and only if they are over a field extension of $K$.

In your situation, you can see your two representatiosn as complex representations, and the corresponding characters will be the same. Hence your representations are isomorphic over the complex numbers, hence over the real numbers.