I was wondering how to prove this. I came across while doing some reading.
Let $G$ be a group and let $N$ be a non-trivial normal subgroup with $C_{G}(N)$ ≤ $N$ and $H_{1}(G/N, Z(N))$ = 1. Then any automorphism of $G$ which fixes $N$ element-wise is inner (on $G$)
Let $\alpha \in {\rm Aut}(G)$ fix $N$ element-wise, let $A= \langle \alpha \rangle \le {\rm Aut}(G)$, and let $X = G \rtimes A$ be the semidirect product, with the natural induced action of $A$ on $G$.
Then, since $C_G(N) \le N$, we have $NC_X(N) = NA \unlhd G$ with $NA \cap G = N$, we have $X/N = G/N \times AN/N$, which means that $\alpha$ induces the identity on $G/N$.
Now $[[A,N],G] = [1,G]=1$ and $[[G,N],A] \le [N,A] = 1$, so by the $3$-subgroups lemma we have $[[A,G],N] = 1$.
Hence, for all $g \in G$, we have $\alpha(g) = \phi(g) g$ with $\phi(g) \in Z(N)$, and you can check that $\phi:G \to Z(N)$ is a derivation. Since $\phi(n)=1$ for all $n \in N$, it induces a derivation $\bar{\phi}:G/N \to Z(N)$.
Then, since $H^1(G/N,Z) = 0$, $\bar{\phi}$ must be a coboundary, which is equivalent to $\alpha$ being equal to conjugation by an element of $Z(N)$. Hence $\alpha$ is inner.