If a right adjoint to the product functor exists, must it be the diagonal? Let $C$ be a category with binary products. The product functor $\times : C^2 \to C$ is right adjoint to the diagonal $\Delta: C \to C^2$. If $C$ has biproducts, then $\times$ is also left adjoint to $\Delta$. But from the fact that $\times$ is left adjoint to some functor $R$, can we conclude that $R = \Delta$?
In the case of nullary products, the answer is yes: if the terminal object $1: C^0 \to C$ has a right adjoint $R$, there's only one functor $C \to C^0$, so of course $R$ coincides with the diagonal $\Delta$. So a terminal object is also initial as soon as the functor $1: C^0 \to C$ has a right adjoint.
To say that $\Delta$ is right adjoint to $\times$ is to say that $\times$ is also a coproduct functor. I believe that if the binary product is also a binary coproduct, this implies that the $C$ is canonically enriched in pointed sets, with the "0" points on the homsets providing the structure maps for the "identity matrix" natural isomorphism $\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}: \amalg \Rightarrow \times$. So in this case $C$ almost has biproducts (except that the 0 object might not actually be representable).
The picture I have in mind is the string of adjoints $\amalg \dashv \Delta \dashv \times$, and the question "when can this adjoint string be extended further?", which leads to the question "if the adjoint string extends further, is it always (as in familiar examples) periodic of period 2?". 
 A: Building on Garlef Wegart's work, the answer is yes in general. The idea is to show that if $C$ has a right adjoint to $\times$, then it is enriched in pointed sets, so that $C$ embeds fully faithfully in a category $C_\ast$ with a zero object and the easy argument can be applied. And as noted already, in this case $C$ will have biproducts with respect to a (necessarily unique) enrichment in pointed sets.
Suppose that $R= (R_1, R_2): C \to C \times C$ is right adjoint to $\times : C \times C \to C$. Then $\pi_1 : X \times Y \to X$ is natural in $X$ and $Y$. By adjointness, this corresponds to $\pi_{11}: X \to R_1 X$ and $\pi_{12}: Y \to R_2 X$ natural in $X$ and $Y$. Similarly, $\pi_2: Y \times X \to X$ gives us $\pi_{21} : Y \to R_1 X$ and $\pi_{22}: X \to R_2 X$. Thus we get $(\pi_{12}, \pi_{21}): Y \to R_2 X \times R_1 X$ natural in $X$ and $Y$, which we can compose (after swapping the factors) with the counit $R_1 X \times R_2 X \to X$ to obtain a map $Y \to X$, natural in $X$ and $Y$.
That is, $C$ admits a (necessarily unique) enrichment in pointed sets. So there is a canonical way to add a zero object, and the resulting inclusion $C \to C_\ast$ is fully faithful. Moreover, $C_\ast$ still has products, computed as in $C$ with $X \times 0 = 0 \times X = X$. One can extend $R$ by hand to a functor $R_\ast: C_\ast \to C_\ast \times C_\ast$ by setting $R_\ast(0) = (0,0)$, and then check by hand that $R_\ast$ is still right adjoint to $\times_{C_\ast}$. (Conceptually, zero objects are absolute for $\mathsf{Set}_\ast$-enrichment, and the forgetful functor from $\mathsf{Set}_\ast$-enriched categories to $\mathsf{Set}$-enriched categories creates products and adjunctions.)
Hence the argument I gave in the comments applies: $X \times Y = (X + 0) \times (0 + Y) = (X \times 0) + (0 \times Y) = X + Y$ in $C_\ast$, because $\times$, as a left adjoint, must preserve colimits (note that the coproduct $(X,0) + (0,Y) = (X,Y)$ holds automatically in $C_\ast$; we need not assume the existence of coproducts). Hence $\times_{C_\ast} = +_{C_\ast}$ and taking adjoints we have $R_\ast = \Delta_{C_\ast}$. By restricting along the fully faithful inclusion $C \to C_\ast$, we get $R= \Delta_C$ as desired.
A: For the first question: Yes.
Let $C$ be a category with finite products and let $\Delta=(\Delta_1,\Delta_2):C\to C\times C$ s.t. $$\times\dashv \Delta.$$
More specifically we find $$[X\times Y, Z]\simeq[(X,Y),\Delta Z]=[X,\Delta_1 Z]\times[Y,\Delta_2 Z].$$ Note that this isomorphism is functorial in all arguments and given $f:X'\to X$ and $g:Y'\to Y$ we find the isomorphism to be compatible with precomposition with $(f,g)$ (I did not check this compatibility with much diligence so please take care here). Now, take $Y=*$ to be the final object (also the neutral object with respect to $\times$). We find $$[X,Z]\simeq[X,\Delta_1 Z]\times [*,\Delta_2 Z].$$
Taking $X=Z$ and we find $\varphi\in [Z,\Delta_1 Z]$ and $\omega\in [*,\Delta_2 Z]$ corresponding to the identity on $Z$. Conversely, taking $X=\Delta_1 Z$ we find $\phi\in[\Delta_1 Z, Z]$ corresponding to $(\mathrm{id}_{\Delta_1 Z},\omega)$. Now the compatibility with precomposition can be used to obtain $$\varphi\circ\phi=\mathrm{id}\textrm{ and } \phi\circ\varphi=\mathrm{id}.$$
I guess you'd still have to confirm that this actually constitutes an isotransformation $\mathrm{id}_C\simeq\Delta_1$.
Edit: After this, I guess we also find $*$ to not only be final but also initial as we arrive at $$(-,\omega):[X,Z]\simeq[X,Z]\times[*,Z].$$ So $[*,Z]=\{\omega\}$.
