Identity about a Functor I'm moving my first steps in CT and suddenly after reading about functors, this question came up in my mind:
Let $F \colon A \to B$ a functor between two categories $A$ and $B$, is it true that for any arrow $f \colon a \to a'$ in $A$, 

is commutative?
According to me the definition of functor doesn't involve nor implies this identity. Nevertheless, due to my restricted experience in CT, I cannot find a counterexample. 
I hope the clarification will help me build a strong "set" of example/counterexample in this field.
Thanks in advance.

In addition to Marco's answer, let me remark that one might construct such an endofunctor with a natural transformation (on a bigger category) for any functor $F:{\bf A}\to{\bf B}$, based on exactly the commutative squares in your question!
So, take the disjoint union of ${\bf A}$ and ${\bf B}$ and take new basic arrows $j_a:a\to F(a)$, and define all compositions  formally so that the commutativities of your squares hold. 
If we want to write it, we can say that the new arrows ('heteromorphisms') $a\to b$ are just pairs $\langle \beta,a\rangle$ where $\beta:F(a)\to b$, and then $j_a=\langle 1_{F(a)},a\rangle$, and composition by an $\alpha:a_0\to a$ is defined as
$$\langle \beta,a\rangle\circ\alpha :=\langle\ \beta\circ F(\alpha),\,a_0\rangle\,.$$
[Note that this is just the (collage of) the profunctor $F_*:{\bf A}^{op}\times{\bf B}\to{\bf Set}$ defined by $(a,b)\mapsto\hom_{\bf B}(F(a),b)$]
