We prove the three conditions are pairwise equivalent by showing (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (i). **Step 1: (i) $\Rightarrow$ (ii).** Assume $gK = Kg$ for all $g \in G$. Then: \begin{align*} gKg^{-1} = \{gkg^{-1} : k \in K\} = (gK)g^{-1} = (Kg)g^{-1} = K(gg^{-1}) = K. \end{align*} **Step 2: (ii) $\Rightarrow$ (iii).** Assume $gKg^{-1} = K$ for all $g \in G$. Then for any $k \in K$ and $g \in G$, the element $gkg^{-1}$ belongs to $gKg^{-1} = K$. So $gkg^{-1} \in K$. **Step 3: (iii) $\Rightarrow$ (i).** Assume $gkg^{-1} \in K$ for all $k \in K$ and $g \in G$. We show $gK \subseteq Kg$ and $Kg \subseteq gK$. For $gK \subseteq Kg$: let $gk \in gK$. By assumption, $gkg^{-1} = k_1$ for some $k_1 \in K$, so $gk = k_1g \in Kg$. For $Kg \subseteq gK$: let $kg \in Kg$. Applying (iii) with $g$ replaced by $g^{-1}$ gives $g^{-1}k(g^{-1})^{-1} = g^{-1}kg \in K$, say $g^{-1}kg = k_2$. Then $kg = gk_2 \in gK$. Therefore $gK = Kg$.