Let $G$ be a discrete group. Let $P$ denote the groupoid whose objects are the elements of $G$, and whose morphisms from $a$ to $b$ are the elements $g\in G$ such that $gag^{-1}=b$. It is then a standard fact that $\Omega'BG=BP$. Less naturally but more concretely, if we choose a set $C$ of representatives for the conjugacy classes in $G$, we get an equivalence $\Omega'BG=\coprod_{c\in C}BZ_G(c)$. If $G$ is nonabelian then this will be different from $BG\times \Omega BG=BG\times G$. For example, you could take $G$ to be the free group on two generators, so $BG$ is a figure eight.
UPDATE: I can't immediately remember a convenient reference for the above "standard fact". However, if we regard $\mathbb{Z}$ and $G$ as groupoids with one object, then there is a functor $e\colon\mathbb{Z}\times P\to G$ given by $e((n,g)\colon a \to b)=g^n$. This gives a map $$ Be \colon S^1\times BP = B(\mathbb{Z}\times P) \to BG, $$ and adjointly a map $BP\to\text{Map}(S^1,BG)=\Omega'BG$, which is the one you want. Another way to analyse this is to consider $S^1$ as the pushout of two semicircles along a copy of $S^0$. After applying the functor $\text{Map}(-,BG)$ (and recalling that the semicircles are contractible) we get a homotopy pullback diagram $$ \begin{array}{ccc} \Omega'BG & \to & BG \\ \downarrow && \downarrow \\ BG & \to & BG\times BG \end{array} $$ After thinking a bit about how the maps work, one can recover the previous description of $\Omega'BG$.