Explain why there is a unique morphism $(\phi\times\phi):G\times G\to H\times H$. 
Let $\phi:G\to H$ be a morphism of category $C$ with products. Explain why there is a unique morphism $$(\phi\times\phi):G\times G\to H\times H$$

How is $G\times G$ defined in a category, where $G$ is an object of the category? 
Motivation: $A\times B$ is the universal object in the category of objects mapping to both $A$ and $B$. Is $G\times G$ the universal object in the category of objects mapping to $G$? If that is the case, wouldn't $\underbrace{G\times G\times\dots\times G}_{\text{n times}}$ be a universal object too?
Also, how is $(\phi\times\phi)$ defined in a category? It seems to me that products are defined only for objects. Is there a category with morphisms as objects that I should consider here?
 A: The idea is this: To define a morphism $f:A \rightarrow B\times C$ into a product is exactly the same as defining morphisms $f_b:A \rightarrow B$ and $f_c:A \rightarrow C$. So you need to come up with a map $\phi_L:G \times G \rightarrow H$ and $\phi_R:G \times G \rightarrow H$. 
One of the other properties of a product $A \times B$ is it comes with projection maps $\pi_L:A \times B \rightarrow A$ and $\pi_R:A \times B\rightarrow B$. That means that $G \times G$ also has these projection maps. Can you see how to put all this information together?
Extra hint: It may or may not be easier for you see what's going on if you take maps $f:A \rightarrow B$ and $g:C \rightarrow D$ and define $f \times g:A \times C \rightarrow B \times D$ and specialize this to your case.
A: Yes, you can also imagine the so called arrow category $C^\to$ of $C$, which has the morphisms of $C$ as objects and commutative squares as morphisms (say from the upper arrow of the square to the lower arrow).
Now, if $\phi:A\to B,\ \psi:C\to D$, then $\phi\times\psi$ will be an arrow $A\times C\to B\times D$. 
In $\mathcal Set$ (and actually in many other categories) it would be the mapping $(a,c)\mapsto (\phi(a),\,\psi(c))$. 
For its categorical construction, see the other answer.
If $(u,v):\alpha\downarrow\phi$ in the arrow category (i.e. $u,v$ are also arrows in $C$ and $v\alpha=\phi u$) $\,$ and $(w,z):\alpha\downarrow\psi$, then use the product property in $C$ for both $A\times C$ and $B\times D$ to define the square $\alpha\to(\phi\times\psi)$ that makes the whole diagram commutative.
