A subgroup is normal exactly when it is closed under conjugacy.Suggest a slogan
Let \(G\) be a group and \(N\leq G\) a subgroup. Then \(N\) is normal in \(G\) if and only if \(N\) is a union of conjugacy classes of \(G\). In that case \(N=\coprod_{i} C(h_i)\), where \(h_i\) runs over a set of representatives of the conjugacy classes of \(G\) contained in \(N\).
If \(N\normal G\) and \(h\in N\), then \(ghg^{-1}\in N\) for all \(g\in G\), so \(C(h)\subseteq N\) and \(N=\bigcup_{h\in N}C(h)\). Conversely, if \(N\) is a union of conjugacy classes, then for \(h\in N\) and \(g\in G\) we have \(ghg^{-1}\in C(h)\subseteq N\), so \(gNg^{-1}\subseteq N\) for all \(g\); applying this to \(g^{-1}\) gives \(gNg^{-1}=N\). Distinct conjugacy classes are disjoint, so the union is disjoint.