For every group G there is a natural group homomorphism G → Aut(G) whose image is the group Inn(G) of inner automorphisms and whose kernel is the center of G. Thus, if G has trivial center it can be embedded into its own automorphism group
The inner automorphism group of a group G, Inn(G), is trivial (i.e., consists only of the identity element) if and only if G is abelian.
Consider group C4 and Homomorpism f: C4->Aut(C4). Inn(C4)={e}. Ker(f)=Center(C4)={all elements of C4}.
But |Aut(C4)|=()=(4)=2, where () is Euler's function. Aut(C4) is isomorphic to C2. There is quotient map from C4 to C2. |Img(f)| = 2
Can you please suggest where is my mistake?
There are two automorphisms of $C_4$, true. But one of them (given by $x\mapsto -x$) isn't an inner / conjugation automorphism, so it is not in the image of $f$. The kernel of $f$ is the center of $C_4$, which is all of it, and the image of $f$ is trivial (it is isomorphic to $C_4/Z(C_4)$, as it should be).
The fact that $\operatorname{Aut}(C_4)$ is (isomorphic to) a quotient of $C_4$ is incidental and not really relevant.