My recent answer to Is it possible to define higher cardinal arithmetics (about defining infinite tetrations) requires something I don't know. Here is the simplest case.
Take a set $S$ and consider $$ \mathbf{Set}\xrightarrow{\hom_{\mathbf{Set}}(\_,S)}\mathbf{Set}^{\mathrm{op}}\xrightarrow{\hom_{\mathbf{Set}}(\_,S)}\mathbf{Set}\xrightarrow{\hom_{\mathbf{Set}}(\_,S)}\mathbf{Set}^{\mathrm{op}}\xrightarrow{\hom_{\mathbf{Set}}(\_,S)}\dots $$ What is the colimit of this diagram?
I realize that formally there is an absolutely explicit construction and description of this colimit. What I want to know is whether the colimit is equivalent to some known manageable category. I don't even see whether it is equivalent to $\mathbf{Set}$.
The same for the (inverse) limit of the backwards diagram $$ \dots\xrightarrow{\hom_{\mathbf{Set}}(\_,S)}\mathbf{Set}^{\mathrm{op}}\xrightarrow{\hom_{\mathbf{Set}}(\_,S)}\mathbf{Set}\xrightarrow{\hom_{\mathbf{Set}}(\_,S)}\mathbf{Set}^{\mathrm{op}}\xrightarrow{\hom_{\mathbf{Set}}(\_,S)}\mathbf{Set} $$
I cannot even figure out whether the limit possesses an initial or a terminal object.
As a (probably) more lightweight version, - replace $\mathbf{Set}$ with the (large) groupoid of sets-and-bijections (but again take the functor $\hom_{\mathbf{Set}}$, rather than $\hom$ in this groupoid (since otherwise it would give something trivial)).
Update
Following suggestion of Philippe Gaucher I checked the simplest cases, with $S$ empty or singleton. In these cases both the limit and colimit are, respectively, $0\to1$ and the terminal category.
But I also realized that both the limit and colimit come out equivalent to their own opposites, so one never gets anything equivalent to $\mathbf{Set}$.
It is thus even more unclear what does one get in the limit for general $S$. Also I still do not see whether initial or terminal object exists there. Already the case of two element $S$ I don't see.