Conjugate subgroups in inverse limit of groups Let $(G_i)_{i\in I}$ be a cofiltered system of finite groups and let $(H_i)_{i\in I}$ such that $H_i\subset G_i$ are subgroups and the $H_i$ form a compatible cofiltered system to the one of the $G_i$, i.e. for all $i,j$ and transitions $i\rightarrow j$, the diagram 
$$\require{AMScd}
\begin{CD}
    H_i @>>> H_j\\
    @V  V V @VV  V\\
    G_i @>>> G_j
\end{CD}$$
commutes, where the vertical arrows are the inclusion maps. Suppose further that every transition map is surjective both for the $G_i$ and the $H_i$.
Now let $(K_i)_{i\in I}$ be another compatible cofiltered system of subgroups of the $G_i$ and suppose everything I assumed for the $H_i$ also holds for the $K_i$. Assume further that for every $i\in I$ there is $g_i\in G_i$ such that $g_iH_ig_i^{-1}=K_i$. Is it true that $\lim_{i\in I} H_i$ and $\lim_{i\in I}K_i$ are conjugate in $\lim_{i\in I}G_i$ as well?
I was thinking about using the fact that the projection maps $\pi_i:\lim_{i\in I}G_i\rightarrow G_i$ are all surjective in order to find an element $g\in\lim_{i\in I}G_i$ such that $\pi_i(g)=g_i$ where the right-hand side is a $g_i$ as specified above. In more detail, my idea is to let $S_i\subset G_i$ be the set of all elements that conjugate $H_i$ and $K_i$ and then take the intersection $\bigcap_{i\in I}\pi_i^{-1}(S_i)$ in $\lim_{i\in I}G_i$. Certainly every element in the intersection will conjugate $\lim_{i\in I}H_i$ and $\lim_{i\in I}K_i$ but I am having trouble proving that this set is non-empty.
Any help is appreciated.
 A: Say the transition maps are $\pi_i^j: G_j\twoheadrightarrow G_i$. Then consider the sets
$$T_A := \{(x_i) \in \prod_{i\in I} G_i \mid \forall i,j\in A: \pi_i^j(x_j)=x_i \wedge x_i H_i x_i^{-1} = K_i\}$$
for all subsets $A\subseteq I$. Since you also know that $\pi_i^j(H_j)=H_i$ and $\pi_i^j(K_j)=K_i$, you can conclude that $T_A\neq\emptyset$ for all finite $A\subseteq I$: Simply choose an $i\in I$ such that $\forall a\in A: i\to a$, set $x_a:=\pi_a^i(g_i)$ and choose $x_j=1$ for all $j\in I\setminus A$. This gives you an element $x\in T_A$.
Now your claim then follows from a compactness argument: Each $G_i$ gets the discrete topology. Then $\prod_{i\in I} G_i$ is compact as a product of compact spaces. The sets $T_A$ are closed w.r.t. to the product topology and satisfy $T_{A_1}\cap T_{A_2}\supseteq T_{A_1\cup A_2}$ by definition. The family $T_A$ with $A$ finite is a family of closed subsets such that all finite intersections of this family are non-empty. By compactness, the intersection of all members in this family is non-empty. That intersection is exactly the set you want.
