Quotient group maps to outer automorphism group of normal subgroup

Statement
Suppose $$G$$ is a group and $$N$$ is a normal subgroup of $$G$$. There is a natural choice of homomorphism of groups from the quotient group $$G/N$$ to the outer automorphism group $$\operatorname{Out}(N)$$:

$$G/N \to \operatorname{Out}(N)$$

defined as follows: for any element of $$G/N$$, pick an element $$g \in G$$ in that coset of $$N$$. Conjugation by $$g$$ induces an automorphism of $$N$$, i.e., an element of the automorphism group $$\operatorname{Aut}(N)$$. Although the automorphism depends on the choice of $$g$$ in the coset of $$N$$, the coset of $$\operatorname{Inn}(N)$$ in $$\operatorname{Aut}(N)$$ for that element is independent of $$g$$.

More explicitly, there is a composite map:

$$G \to \operatorname{Inn}(G) \to \operatorname{Aut}(N)$$

Under this map, the image of the subgroup $$N$$ of $$G$$ lies inside $$\operatorname{Inn}(N)$$. Thus, we get an induced map from the quotient group $$G/N$$ to the quotient group $$\operatorname{Aut}(N)/\operatorname{Inn}(N) = \operatorname{Out}(N)$$.