Limit of Hom-groupoid 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
$$\underline{\text{Hom}}(G,\lim_{i\in I} H_i)\cong \lim_{i\in I}\underline{\text{Hom}}(G, H_i)?$$
 A: 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 $\text{Grpd}(X^{\delta}, BG) \cong * \cong \text{Grpd}(*, BG)$.
