"Natural" in the context of category theory means natural transformation. A "natural isomorphism" is just an isomorphism between functors, i.e. mutually inverse natural transformations. Sometimes we say something like "$\text{Hom}(FA,B) \cong \text{Hom}(A,UB)$ is natural in $A$ and $B$" by which we mean $(A,B)\mapsto\text{Hom}(FA,B)$ and $(A,B)\mapsto\text{Hom}(A,UB)$ are isomorphic as functors. While natural transformations are indeed often "natural" in the non-technical sense, there are a lot of them, and they can be quite arbitrary. For example, in the category of commutative monoids (or groups), $x \mapsto 42x$ is a natural transformation.
"Canonical" doesn't have a widely used technical meaning as far as I'm aware. (It does have some technical definitions, but the ones I'm aware of are not widely used.) The categorical notion that unambiguously comes closest to capturing the notion of a "canonical map" is the notion of a universal arrow. I'll use a variation on the presentation using a notion called a universal element. I describe the relation between these notions here. We say a functor $F$ is representable if $\text{Hom}(X,-) \cong F$ for a given object $X$. (Note, that in words this says $F$ is naturally isomorphic to a hom-functor.) We'll equivalently say that $X$ represents $F$. A universal element is simply the image of $id$ under this isomorphism, i.e. if $\varphi:\text{Hom}(X,-)\cong F$, then $\varphi(id)\in F(X)$ is the universal element. In the (very common, in fact necessary) case that $F$ itself is some sort of hom-functor, then the universal element will be an arrow in some category and that arrow is the universal arrow. An excellent exercise is to show that if $X'$ also represents $F$ then $X \cong X'$ and that isomorphism is unique.
The two examples of "canonical maps" from Omnomnomnom's second link are of this form. (Note, the other two examples are examples of "structure maps".) This may not be obvious, so let me spell it out.
The first example is the quotient group construction. Let $N$ be a normal subgroup of a group $G$. Define $F(H)\equiv\{f\in\text{Hom}(G,H)\mid \forall x\in N.f(x) = 0\}$. Then it is the case that $\text{Hom}(G/N,-)\cong F$. (Prove this. Don't forget naturality.) The universal element is a group homomorphism $q\in\text{Hom}(G,G/N)=F(G/N)$ for which $q(x)=0$ iff $x\in N$. In particular, any $f \in F(H)$ factors through $q$ uniquely.
The second example says there is a natural transformation $\lambda : V \to V^{**}$ natural in $V$ where $V$ is a vector space over the field $k$, and $V^* \equiv V \multimap k$ is the duality with $V\multimap W$ being the vector space of linear functions from $V$ to $W$. We have the following adjunction $$\text{Hom}(U,V\multimap W)\cong\text{Hom}(V,U\multimap W)$$ natural in $U$, $V$, and $W$. (Prove this.) Set $F_V(U)\equiv\text{Hom}(V,U\multimap k)=\text{Hom}(V,U^*)$. $\text{Hom}(-,V^*)\cong F_V$. (This looks slightly different from our definition of representability, but it's actually the same. Why?) The universal element is then a map $\lambda_V \in F_V(V^*) = \text{Hom}(V,V^{**})$. These arrows indexed by $V$ form the components of a natural transformation $\lambda : Id \to (-)^{**}$. In fact, this result can be proved generally, and the general result is a corollary to a result called parameterized representability.
