The answer is no. You can easily have a situation where $f$ is the trivial map, while $g$ makes $H$ act through an inner automorphism of $N$, so that in both cases $N\rtimes H\cong N\times H$. For concreteness, let $N=D_8$ be the dihedral group of order 8, let $\sigma$ be a non-central involution in $N$, let $H=\langle h\rangle\cong C_2$, and define $g(h)(n)=\sigma^{-1} n\sigma\;\forall n\in N$. Then $G=N\rtimes_g H\cong N\times H$, since the subgroup $\langle h\sigma\rangle$ of $G$ is of order 2, intersects $N$ trivially, and commutes with it.