Simplicial nerve functor commutes with opposites 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?
 A: (This is a supplement to Harry's answer which doesn't quite fit into a comment.)
As Harry mentions, it suffices to show that the cosimplicial objects $\mathfrak{C}((\Delta^\bullet)^\mathrm{op})$ and $(\mathfrak{C}(\Delta^\bullet))^\mathrm{op}$ are isomorphic. This follows from the following observation: Although the simplicial category $\mathfrak{C}(S)$ is rather mysterious for a simplicial set $S$, in the case $S$ is the nerve of a poset, it admits a completely explicit description analogous to the definition of $\mathfrak{C}(\Delta^n).$ Namely, if $P$ is a poset, then $\mathfrak{C}(N(P))$ is the simplicial category whose objects are $P$, and whose hom-simplicial sets are given by
$$\mathfrak{C}(N(P))(x,y)=N(\{I\subset P \mid I\text { a finite totally ordered set, }\min I=x,\,\max I=y\}).$$
The composition is induced by the operation of union. One can verify easily that $\mathfrak{C}(N(P))$ is as stated above by directly verifying the universal property of $\operatorname{colim}_{\Delta^n\downarrow N(P)}\mathfrak{C}(\Delta^n)$.
With this description, we have a natural (in $P$) isomorphism $\mathfrak{C}(N(P^\mathrm{op}))\cong\mathfrak{C}(N(P)^\mathrm{op})=\mathfrak{C}(N(P))^\mathrm{op}$. This makes the claim at the beginning transparent.
A: 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.
