It might be a trivial question. So, I apologise in advance. Let $ \Delta ^{op}_n $ be the full subcategory of $ \Delta ^{op} $ such that the set of objects of $ \Delta ^{op}_n $ is $ \left\{ 0, \ldots , n\right\} $. The inclusion $ i: \Delta ^{op} _ n \to \Delta ^{op} $ induces a functor $ i ^*: sSet\to Fun (\Delta _n ^{op}, Set ) $. I wonder if this functor has a right adjoint.
On the other hand, I guess that this functor is right adjoint and the left adjoint is the functor which associates each truncated simplicial set to the "degenerated extension" (by degenerated extension, I mean that the higher simplices are the degenerated k-simplices for $ k\leq n $). Am I right?
By general abstract nonsense, given any functor $F:\mathcal C \to \mathcal D$, the induced functor $F^*:\hat{\mathcal D} \to \hat{\mathcal C}$ between the associated presheaf categories has both a left adjoint $F_!$ and a right adjoint $F_*$ (the left and right Kan extensions).
For your case above, the left/right adjoint will map an $n$-truncated simplicial set $X$ to a simplicial set whose $\le n$ simplices agree with those of $X$, and the other simplices are all degenerate. There are two extreme choices for the new generate simplices: add as many as possible, or add as few as possible. The left adjoint adds as many as possible, and (as you claim, I think) is obtained by "pulling back" the lower degree simplicies (in fact, it suffices to only pullback the degree $n$ simplices) along the structure maps. The right adjoint adds just one new simplex at dimensions $>n$.