Relationships between initial/terminal objects and initial/terminal morphisms (if any) in the same category. The definition of initial and terminal objects given here http://en.wikipedia.org/wiki/Initial_and_terminal_objects makes sense to me.
The definition of initial and terminal morphisms given here http://en.wikipedia.org/wiki/Universal_property#Initial_morphism makes sense to me.
Now, suppose $C$ some category and $U:C\rightarrow C$ is some endofunctor.
What relationship, if any, is there between the initial objects of $C$ and the initial objects of $(X\downarrow U)$ for some object $X$ in $C.$
 A: Martin Brandenburg has commented that an initial morphism is an initial object in a certain category. I want to point out that conversely, an initial object can be viewed as a special case of an initial morphism. 
Let $D$ be any category. Let $C$ be the terminal category consisting of a single object $X$ and a single arrow $\text{id}_X:X\to X$. Let $U:D\to C$ be the unique functor sending every object to $X$ and every arrow to $\text{id}_X$. Then an initial morphism from $X$ to $U$ is defined by an object in $D$, call it $A$, and a morphism in $C$ from $X$ to $U(A) = X$ (the only choice is $\text{id}_X$). 
Now the definition reads that for any object $Y$ in $D$ and any morphism $f:X\to U(Y)$ in $C$, there is a unique morphism $g:A \to Y$ in $D$ such that a diagram commutes: $U(g)\circ \text{id}_X = f$ in $C$. But the choice of the morphism $f$ and the commutativity of the diagram are trivial, so this boils down to the statement that for any $Y$ in $D$, there is a unique morphism $g:A\to Y$, i.e. $A$ is an initial object in $D$.
Another way of saying this is that the functor $A:C \to D$ which sends the unique object in $C$ to the initial object in $D$ is left adjoint to the trivial functor $U:D\to C$. Note that $\text{Hom}_C(X,U(Y))$ consists of exactly one element ($\text{id}_X$) for all $Y$, and $\text{Hom}_D(A(X),Y)$ consists of exactly one element for all $Y$ if and only if $A(X)$ is initial.
Oh, you've just changed your question, so my answer no longer applies. Still, you may find it helpful.
