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?". 



If a right adjoint to the product functor exists, must it be the diagonal?
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 $[\*,Z]=\{\omega\}$.


