There are two "opposite" functors: 
$$ op\_\Delta\colon sSet\to sSet$$
and
$$op\_s\colon sCat\to sCat.$$
The first takes a simplicial set to its opposite simplicial set by precomposing with the opposite of a functor $\Delta\to \Delta$ which is the identity on objects and takes a morphism $\langle k\_0,\ldots,k\_n\rangle\colon [n]\to [m]$ (where $k\_i$ is the integer that $i$ gets mapped to by this morphism) to the morphism $\langle m-k\_n,\ldots,m-k\_0\rangle$. For example, the morphism $[1]\to [2]$ that takes $0$ to $0$ and $1$ to $1$ gets mapped to the morphism that takes $0$ to $1$ and $1$ to $2$. 


The second functor takes a simplicial category to the opposite simplicial category, which is easier to define. It has the same objects but given $x,y\in C^{op}$, the mapping complex $C^{op}(x,y)=C(y,x)$. 


There is also the simplicial nerve functor $N\colon sCat\to sSet$. I am interested in a proof of the fact that for a given fibrant simplicial category $C$, there is a weak equivalence of quasicategories $op\_\Delta\circ N(C)\simeq N\circ op\_s(C)$.


I'm relatively certain that this is an elementary proof, but I don't feel skilled enough with the simplicial nerve to figure out the details. Does anyone have a proof of this fact?



Simplicial nerve functor commutes with opposites
It all follows from the following elementary lemma: 


$\mathfrak{C}([n]^{op})$ is isomorphic to $\mathfrak{C}([n])^{op}$ as a cosimplicial simplicial category (in fact, they are actually equal, since the components of the natural isomorphism are all identities). 


proof: It is an immediate calculation from the definition of opposites. 


Consequence: From this lemma, apply adjunctions to show that $^{op}$ actually commutes with the simplicial nerve up to isomorphism. I can fill in details if you want.


