For any groups $G,H$, we can define the category, in fact a groupoid,
$$\underline{\text{Hom}}(G,H)$$
whose objects are group morphisms $G\to H$ and morphisms $(f:G\to H)\to (g:G\to H)$ are elements of $H$ conjugating $f$ into $g$. Let $\{H\_i\}\_{i\in I}$ be an inverse system of groups. Then of course we have $\text{Hom}(G,\lim\_{i\in I} H\_i)\cong \lim\_{i\in I}\text{Hom}(G, H\_i)$. My question is do we also have
$$\underline{\text{Hom}}(G,\lim\_{i\in I} H\_i)\cong \lim\_{i\in I}\underline{\text{Hom}}(G, H\_i)?$$



Limit of Hom-groupoid
Sure, this follows from the Yoneda lemma. Let $B : \text{Grp} \to \text{Grpd}$ be the functor sending any group $G$ to the groupoid with one element $\star$ such that $BG(\star, \star) = G$. For any groupoid $X$ we have
\begin{align}
\text{Grpd}(X, \underline{\text{Hom}}(G,\lim\_{i\in I} H\_i))
&\cong \text{Grpd}(X, \underline{\text{Grpd}}(BG, B\lim\_{i\in I} H\_i))
\\&\cong \text{Grpd}(X \times BG, B\lim\_{i\in I} H\_i)
\\&\cong \lim\_{i \in I}\text{Grpd}(X \times BG, BH\_i)
\\&\cong \lim\_{i \in I}\text{Grpd}(X, \underline{\text{Grpd}}(BG, BH\_i))
\\&\cong \lim\_{i \in I}\text{Grpd}(X, \underline{\text{Hom}}(G, H\_i))
\\&\cong \text{Grpd}(X, \lim\_{i \in I}\underline{\text{Hom}}(G, H\_i))
\end{align}
using the universal property of the internal hom and the continuity of $B$ and the Yoneda embedding. So by the fully-faithfulness of the Yoneda embedding we have the isomorphism in your question.


To show that $B$ preserves limits it's not too hard to show directly that it preserves products and equalizers. But it also preserves limits because it's a right adjoint. Its left adjoint can be constructed as follows. Let $\text{Grpd}\_{1}$ be the full subcategory of $\text{Grpd}$ on the one-element groupoids. Then $B : \text{Grp} \to \text{Grpd}\_{1}$ is an equivalence, so it suffices to give a left adjoint to the inclusion $\text{Grpd}\_{1} \to \text{Grpd}$, which I'll also abusively call $B$.


For any groupoid $X$, let $X^{\delta}$ be the discrete groupoid on the underlying set of $X$, and let $i\_{X} : X^{\delta} \to X$ be the canonical inclusion. The pushout of $i\_{X}$ along the unique functor $X^{\delta} \to \*$ gives a one-element groupoid $LX$ (it's not hard to show that this has one element, using the fact that the underlying set functor $\text{Grpd} \to \text{Set}$ is a left adjoint). For any one-element groupoid $G$ we therefore have
\begin{align}
\text{Grpd}\_{1}(LX, G)
&=
\text{Grpd}(LX, G)
\\&\cong
\text{Grpd}(\*, BG)
\times\_{\text{Grpd}(X^{\delta}, BG)}
\text{Grpd}(X, BG)
\\&\cong
\text{Grpd}(X, BG)
\end{align}
by the definition of $\text{Grpd}\_{1}$, the universal property of $LX$, and the isomorphism $\text{Grpd}(X^{\delta}, BG) \cong \* \cong \text{Grpd}(\*, BG)$.


