 Solution

Take $g=(12\cdots n)$. Since $n\equiv 3(\mathrm{mod}4)$ is odd, its sign is $(-1{)}^{n-1}=1$, so $g\in A_n$. Define $s$ to fix $1$ and reverse the other cyclic labels: it interchanges $2$ with $n$, $3$ with $n-1$, and so on. Then $sg{s}^{-1}=g^{-1}$, whileThe centralizer of an $n$-cycle in $S_n$ is $⟨g⟩$: a commuting permutation is determined by the image of one point, and that image fixes the corresponding power of the cycle. All its elements are even, because $g$ is even. Any other conjugator taking $g$ to $g^{-1}$ differs from $s$ by a centralizer element and therefore remains odd. Thus $g$ and $g^{-1}$ are not conjugate in $A_n$. This is the parity obstruction described by inversion of an odd cycle in an alternating group.

For any complex character of a finite group, a unitary realization givesIf every $\theta \in \mathrm{Irr}(A_n)$ had real value at $g$, all these characters would take equal values on $g$ and $g^{-1}$. But irreducible characters separate conjugacy classes: they form a basis of class functions, so equality of every irreducible value would also give equality of the class indicator functions. The two distinct conjugacy classes cannot have that property. ConsequentlyIn other words, these alternating groups are not ambivalent groups, even though all symmetric-group irreducible characters are integer-valued.