If $N,K$ are two finite groups with isomorphic subgroups, show $ X \leq N \times K $ is normal.

If $N,K$ are finite groups with normal subgroups $N_{1} \mathrel{\unlhd}N,\ K_{1} \mathrel{\unlhd}K $ such that $ \varphi : N_{1} \rightarrow K_{1} $ is an isomorphism, then show $X := \left\{ (n_{1}, \varphi(n_{1})) | n_{1} \in N_{1} \right\}$ is normal in $N \times K$.

I proceeded with the natural way of proving normality. Suppose $(n,k) \in N \times K $ and $(n_{1}, \varphi(n_{1})) \in X$. Then:

$(n,k)^{-1}(n_{1}, \varphi(n_{1}))(n,k) $
$=(n^{-1},k^{-1})(n_{1}, \varphi(n_{1}))(n,k)$
$=(n^{-1}n_{1}n,\;k^{-1}\varphi(n_{1})k \ $ )

By normality of $N_{1}, K_{1}$ in $N$ and $K$ respectively it is the case that $ n^{-1}n_{1}n \in N_{1} \:$ and $k^{-1}\varphi(n_{1})k \ \in K_{1}$.

To finish, one must show $\varphi(n^{-1}n_{1}n) = k^{-1}\varphi(n_{1})k$. But this is where I am stuck. As $n$ need not be in $N_{1}$, $\varphi(n)$ may not be defined. There is no relation between $n,k$, so I am unsure how to proceed.

This is not true.

If $N=K=N_1=K_1=S_3,$ and $\phi$ is the identity, this amounts to $X=\{(g,g)\mid g\in S_3\}.$

Show $X$ is not a normal subgroup of $S_3\times S_3.$

For example: Try $h=((12),e)\in S_3\times S_3$ and $g=((123),(123))\in X.$ Show $hgh^{-1}\notin X.$