= Solution
Take $g=(1\ 2\ \cdots\ n)$. Since $n\equiv3\pmod4$ 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 $sgs^{-1}=g^{-1}$, while
$$
\operatorname{sgn}(s)=(-1)^{(n-1)/2}=-1.
$$
The <centralizer> of an $n$-cycle in $S_n$ is $\langle g\rangle$: 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 gives
$$
\theta(g^{-1})=\overline{\theta(g)}.
$$
If every $\theta\in\operatorname{Irr}(A_n)$ 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. Consequently
$$
\boxed{\text{There is }\theta\in\operatorname{Irr}(A_n)\text{ with }\theta((1\ 2\ \cdots\ n))\notin\mathbb R.}
$$
In other words, these alternating groups are not <ambivalent groups>, even though all symmetric-group <irreducible characters> are integer-valued.
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