How to prove two categories are isomorphic? I am working thru category theory on my own (so bear with me).  What I am looking for is one or two "standard" proofs that two categories are isomorphic.  In other words, I need a couple of boiler-plate proofs that demonstrate the general methodology, preferably involving categories that one frequently encounters.
I of course know that any proof is really just a demonstration that the two categories in question fit the definition (namely that for two categories $C$ and $D$, there exists two functors $ F \colon C \to D $ and $ G \colon D \to C $ such that $ G \circ F $ equals the identity functor on $C$, and $ F \circ G $ equals the identity functor on $ D $.). But how to prove conformity to the definition? Needless to say, any demonstration on a subset of either caregory is no proof.  Similarly, any conjecture that "we've covered all the cases" can certainly be wrong, categories often being so large and complex.  (For me, this problem is particularly acute when dealing with certain dual categories where some morphisms don't really make sense in a normal mathematical way.)
The only proof I can see is a demonstration that lack of isomorphism leads to a contradiction.  Here I need some help.  Preferably "nice" examples.  Thanks.
 A: 
Needless to say, any demonstration on a subset of either caregory is no proof. Similarly, any conjecture that "we've covered all the cases" can certainly be wrong, categories often being so large and complex.

If you show that the composition $FG$ is the identity on a generic object and morphism, and vice versa, you are done. You just need to use the general properties in a smart way, just like in any other branch of mathematics. You just need to show it for a single object and a single arrow if they are general enough. 

A trivial example: consider a set $S$ and let $\mathbb C$ be the category of subsets of $S$ with maps being inclusions. It has a terminal object $S$ and initial object $\varnothing$. 
(The importance of this is the following: an initial object is terminal in the opposite category. If a category has an initial object but no terminal object it can't be isomorphic or equivalent to it's dual.)
You can construct an isomorphism $F:\mathbb C\to \mathbb C^\text{op}$ easily by sending


*

*every subset $X\subset S$ to its complement $FX=S\smallsetminus X$ and 

*every inclusion $i:X\to Y$ to the "reverse inclusion"
$Fi:(S\smallsetminus X) \to (S\smallsetminus Y)$.
I believe that finding the inverse $G$ is quite easy.
Now


*

*you need to verify that given any $X\in \mathbb C$, we have
that $GFX=X$.

*you need to verify that any inclusion $X\to Y$ is mapped to
itself. 

*you need to repeat the verifications for the other composition $FG$.
A: Regard a field $\mathbb{F}$ as a (pre)additive* category with a single object and call this category $\mathbb{F}$ again.  The category of all additive functors $F: \mathbb{F} \rightarrow \mathbf{AbGps}$ is isomorphic to the category of $\mathbb{F}$-vector spaces.  
The other good example which is pretty well known (you see it in the first week of your first course in repn theory) is that the category of $\mathbb{F}G$-modules is isomorphic to the category of representations of the group $G$ over the field $\mathbb{F}$.
*Terminology differs, hence the (pre).  You'll have to figure it out.

OK, as requested, a simpler example.  You may find this example to be incredibly superficial once you understand it.  That's why the most interesting equivalences are natural equivs, not isos.
For an integer $n \geq 0$ let $\mathbf{n} = \{1, 2, \ldots, n\}$ where $\mathbf{0}=\varnothing$.  Let $\mathbb{U}$ be the category of all set-functions on all the objects $\mathbf{n}$.  So, a morphism $f: \mathbf{m} \rightarrow \mathbf{n}$ is just an ordinary, run of the mill function.
(I am using $\mathbb{U}$ for the ``unbased" category.)
For the same integers, let $\mathbf{n}_+ =\{0, 1, \ldots, n\}$ where we will treat $0$ as a basepoint.  Let $\mathbb{B}$ be the category with objects $\mathbf{n}_+$ and morphisms $f: \mathbf{m}_+ \rightarrow \mathbf{n}_+$ ordinary set functions with the property that $f(0)=0$.  (I am using $\mathbb{B}$ for the ``base-pointed" category.)
Finally, let $\mathbb{A}$ be the subcategory of $\mathbb{B}$ consisting of all the objects but only the maps $f: \mathbf{m}_+ \rightarrow \mathbf{n}_+$ which satsify
$$
f(x) = 0 \Longleftrightarrow x=0.
$$
You can check that $\mathbb{A}$ is a legit subcategory.
Some "good" exercises:


*

*$\mathbb{U}$ is naturally equivalent to the category of all finite sets.  It is, however, not isomorphic to this category.  (It is something called a "skeleton.")

*$\mathbb{U}$ and $\mathbb{B}$ are neither iso nor equivalent.

*$\mathbb{U}$ and $\mathbb{A}$ are isomorphic by either appending or forgetting the basepoint.  (This is what you actually requested.)

A: Here is another example, that again, is so "lame" I hesitate to post it. Let $\mathcal{D}$ be the category with objects $(G,z)$ where $G$ is a group and $z: G \to \{e\}$ is a (the) trivial homomorphism; and with arrows $[f]:(G,z) \to (G',z')$ where $f:G \to G'$ is any homomorphism such that $z'f = z$ (which is, of course, any homomorphism). So, strictly speaking, arrows are commutative triangle diagrams "over $\{e\}$", and objects are group zero-homomorphisms. 
Let $\mathcal{C} = \mathbf{Grp}$ with objects that are groups, and arrows that are the group homomorphisms.
Define $F:\mathcal{C} \to \mathcal{D}$ by $FG = (G,z)$, and for $h:G \to H$, $Fh = [h]$ (why does this make sense?).
If $FG = FH$, then $(G,z) = (H,z')$, whence $G = H$, and $F$ is injective on objects.
On the other hand, for any object $(G,z)$ of $\mathcal{D}$, we clearly have $FG = (G,z)$, that is, $F$ is bijective on objects.
In a similar fashion, $F$ is likewise bijective on arrows. We could explicitly construct an inverse $G$ to $F$, but we don't have to.
Can you see how this depends on the uniqueness of the trivial group homomorphism?
