[proofplan] We prove the equivalence in both directions. If $A$ is conjugation-invariant, then every element of $A$ brings its entire [conjugacy class](/page/Conjugacy%20Class) with it, so $A$ is exactly the union of the conjugacy classes indexed by elements of $A$. Conversely, if $A$ is a union of conjugacy classes, then conjugating an element of $A$ keeps it inside the same conjugacy class, hence inside the union. [/proofplan]

[step:Show that conjugation-invariance contains every conjugacy class through a point of $A$]Assume that $A$ is conjugation-invariant. Define the indexing subset $S \subset G$ by $S := A$. We prove \begin{align*} A = \bigcup_{a \in A} \operatorname{Cl}_G(a). \end{align*} First let $x \in A$. Since the identity element $1_G \in G$ satisfies $1_G x 1_G^{-1} = x$, we have $x \in \operatorname{Cl}_G(x)$. Because $x \in A$, this gives \begin{align*} x \in \bigcup_{a \in A} \operatorname{Cl}_G(a). \end{align*} Thus \begin{align*} A \subset \bigcup_{a \in A} \operatorname{Cl}_G(a). \end{align*} Conversely, let $x \in \bigcup_{a \in A} \operatorname{Cl}_G(a)$. Then there exist $a \in A$ and $h \in G$ such that $x = hah^{-1}$. Since $A$ is conjugation-invariant and $a \in A$, it follows that $hah^{-1} \in A$. Hence $x \in A$, so \begin{align*} \bigcup_{a \in A} \operatorname{Cl}_G(a) \subset A. \end{align*} The two inclusions prove \begin{align*} A = \bigcup_{a \in A} \operatorname{Cl}_G(a), \end{align*} so $A$ is a union of conjugacy classes of $G$.[/step]

[guided]Assume that $A$ is conjugation-invariant, meaning that whenever $a \in A$ and $h \in G$, the conjugate $hah^{-1}$ also lies in $A$. We want to express $A$ as a union of conjugacy classes. The natural candidate is the union of all conjugacy classes passing through elements already in $A$, so define the indexing subset $S \subset G$ by $S := A$. We prove the equality \begin{align*} A = \bigcup_{a \in A} \operatorname{Cl}_G(a) \end{align*} by proving both inclusions. First let $x \in A$. The conjugacy class of $x$ contains $x$ itself, because the identity element $1_G \in G$ gives \begin{align*} 1_G x 1_G^{-1} = x. \end{align*} Therefore $x \in \operatorname{Cl}_G(x)$. Since $x \in A$, the class $\operatorname{Cl}_G(x)$ is one of the sets appearing in the union indexed by $a \in A$. Hence \begin{align*} x \in \bigcup_{a \in A} \operatorname{Cl}_G(a). \end{align*} This proves \begin{align*} A \subset \bigcup_{a \in A} \operatorname{Cl}_G(a). \end{align*} For the reverse inclusion, let \begin{align*} x \in \bigcup_{a \in A} \operatorname{Cl}_G(a). \end{align*} By the definition of union, there is some $a \in A$ such that $x \in \operatorname{Cl}_G(a)$. By the definition of conjugacy class, this means that there exists $h \in G$ with \begin{align*} x = hah^{-1}. \end{align*} Now we use conjugation-invariance of $A$: since $a \in A$ and $h \in G$, the element $hah^{-1}$ belongs to $A$. Therefore $x \in A$. This proves \begin{align*} \bigcup_{a \in A} \operatorname{Cl}_G(a) \subset A. \end{align*} Combining the two inclusions yields \begin{align*} A = \bigcup_{a \in A} \operatorname{Cl}_G(a). \end{align*} Thus $A$ is a union of conjugacy classes of $G$.[/guided]

[step:Show that a union of conjugacy classes is closed under conjugation] Assume that there exists a subset $S \subset G$ such that \begin{align*} A = \bigcup_{s \in S} \operatorname{Cl}_G(s). \end{align*} Let $a \in A$ and let $h \in G$. Since $a \in A$, there exists $s \in S$ such that $a \in \operatorname{Cl}_G(s)$. Hence there exists $k \in G$ such that \begin{align*} a = ksk^{-1}. \end{align*} Then \begin{align*} hah^{-1} &= h(ksk^{-1})h^{-1} \\ &= (hk)s(hk)^{-1}. \end{align*} Because $hk \in G$, this shows that $hah^{-1} \in \operatorname{Cl}_G(s)$. Since $s \in S$, we have $\operatorname{Cl}_G(s) \subset A$, and therefore $hah^{-1} \in A$. Thus, for every $a \in A$ and every $h \in G$, one has $hah^{-1} \in A$. Hence $A$ is conjugation-invariant. [/step]