Let $\mathcal{C}$ be a symmetric closed monoidal category. This means in particular that for every fixed object $B \in \mathcal{C}$ the functor $-\otimes B$ has a right adjoint $[B,-]$, i.e. for all $A,B,C$ we have an isomorphism $$ \Phi: \mathcal{C}(A\otimes B,C) \overset{\cong}\longrightarrow \mathcal{C}(A,[B,C]), $$ natural in $A$ and $C$.
It is now claimed here https://ncatlab.org/nlab/show/closed+monoidal+category under '2.Definitions - Symmetric closed monoidal category' that this isomorphism is natural in all three arguments.
On first sight it seems this should follow easily by somehow involving the symmetry on the left side but i am not able to proof it.
Does this even follow from the outset or do we have to assume it additionally?
EDIT:
For my question to even make sense we would first have to assemble all the functors $[B,-]$ into a functor $[-,-]:\mathcal{C}^{op}\times \mathcal{C} \to \mathcal{C}$.
It turns out that this can be done in a unique way such that $\Phi$ is then natural in all three arguments.
To see this let $f:B\to B'$ in $\mathcal{C}^{op}$ (i.e. $f:B'\to B$ in $\mathcal{C}$) and $g:C \to C'$. Furthermore let $Y$ denote the Yoneda embedding $\mathcal{C} \to \mathcal{C}^\wedge$ and $Y^{-1}$ its inverse defined on the image of $Y$.
We have a natural transformation $$f^*g_* : \mathcal{C}(-\otimes B,C) \to \mathcal{C}(-\otimes B',C')$$ and by conjugation with $\Phi$ we obtain $$ \Phi f^*g_* \Phi^{-1}: [B,C]^\wedge \to [B',C']^\wedge.$$ Applying $Y^{-1}$ gives a morphism $[B,C] \overset{[f,g]}\longrightarrow [B',C']$ and this clearly defines a functor $[-,-]:\mathcal{C}^{op}\times \mathcal{C} \to \mathcal{C}$.
By construction of $[-,-]$, $\Phi$ is now natural in all arguments. Furthermore, fixing the first argument of $[-,-]$ gives back the functors we started with.
I guess this answers my original question. But now a new one arises. If we already start with a bifunctor $[-,-]: \mathcal{C}^{op} \times \mathcal{C} \to \mathcal{C}$ such that for every fixed $B$, $[B,-]$ is right adjoint to $-\otimes B$ does it then follow that it is isomorphic to the one constructed above? This is equivalent to asking if $\Phi$ is natural in all arguments in this case.
This is unrelated to symmetry, being true for every monoidal closed category, and isn't an additional assumption. Instead, it is a special case of a result on "adjunctions with a parameter":
Let $F : \mathcal A × \mathcal B → \mathcal C$ be a bifunctor, and denote with $F_B : \mathcal A → \mathcal C$ the functor $F(-, B)$ for every $B ∈ \mathcal B$. If every $F_B$ is left adjoint to a $G_B : \mathcal C → \mathcal A$ via $$φ^B_{AC} : \mathrm{Hom}(F_BA, C) ≅ \mathrm{Hom}(A, G_BC),$$ then there exists a unique functor $G : \mathcal B^{\mathrm{op}} × \mathcal C → \mathcal A$ such that $G_B = G(B, -)$, and $φ$ is natural in $B$ too.
You can find the result and a proof in Mac Lane's Categories for the Working Mathematician.
So as you can see, $[B, -]$'s assemble into a single bifunctor automatically, but note that this functor ("internal hom") will be contravariant in its first argument, just like the usual hom functor is.