Let, $N,K,N',K'$ be groups and $N\cong N',K\cong K'$. Does $N\rtimes K\cong N'\rtimes K'$?

Question

I have tried this problem in the following way-
Let $f:N\to N'$ and $g:K\to K'$ be two isomorphisms. $K, K'$ act on $N,N'$ respectively such that $k•(n_1n_2)=(k•n_1)(k•n_2)\ \forall k\in K, n_1,n_2 \in N$ and $k'•(n_1'n_2')=(k'•n_1')(k'•n_2')\ \forall k'\in K', n_1',n_2' \in N'$.
That give rise to two groups $N\rtimes K$ and $N'\rtimes K'$.
I define $h:N\rtimes K\to N'\rtimes K'$ by $h(n,k)=(f(n),g(k))$
Now I tried to show this $h$ is homomorphism.
$h((n_1,k_1)(n_2,k_2))=h(n_1(k_1•n_2),k_1k_2)=(f(n_1(k_1•n_2)),g(k_1k_2))=(f(n_1)f(k_1•n_2),g(k_1)g(k_2))$
Now, $h(n_1,k_1)h(n_2,k_2)= (f(n_1),g(k_1))(f(n_2),g(k_2))= (f(n_1)(g(k_1)•f(n_2)),g(k_1)g(k_2))$
Now, $h((n_1,k_1)(n_2,k_2))=h(n_1,k_1)h(n_2,k_2)$ iff $f(k_1•n_2)=g(k_1)•f(n_2)$.
I am unable to show this. I think this is not true. Do I need to change my map $h$?