Context: A finite group has all characters real valued if and only if every element $ g \in G $ is in the same conjugacy class as $ g^{-1} $. This property has some special name but I can't remember what it is.
Question: Let $ G $ be a group with this property (that all characters are real valued). Let $ \pi $ be a unitary representation of $ G $. Let $ * $ denote complex conjugation. Is it the case that $ \pi(g)^* \in \pi(G) $ for all $ g \in G $?
Now suppose that $ \pi $ is faithful. And let $ \sigma $ be an outer automorphism of $ G $. Is it the case that $$ \pi(\sigma(g))^*=\sigma (\pi(g)^*) $$ for all $ g \in G $? (If you want you can reinterpret the $ \sigma $ on the right hand side as $ \pi \sigma \pi^{-1} $ )
Work so far: If the Frobenius-Schur indicator of every character of $ G $ is +1 (so every irrep of $ G $ is real) then this is obviously true since $ \pi(g) $ will always be an orthogonal matrix. However there are many groups for which $ g $ is always conjugate to $ g^{-1} $ even though not all irreps are real. An example of such a group is $ SL(2,5) $. I checked degree 2 cases for $ SL(2,5) $ and it seems true so far.
Here is a counterexample it sounds like you aren't expecting. Let $G = \{\pm1\} \cong C_2$ and define $\pi : G \to U(2)$ by $$\pi(-1) = \begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}.$$ Then $\pi(-1)^* \ne \pi(-1)$, so $\pi(-1)^* \notin \pi(G)$.