Generator of Normal Group Let $G$ be a group and $S,T \subset G$ are subsets of $G$  such that $\langle T\rangle =G$ and $\langle S \rangle=N$. Does the following hold:

$N$ is normal in $G$ iff for all $t\in T$ we have $tSt^{-1} \subseteq N$

Here $\langle X \rangle$ denotes the subgroup generated by $X$ (the smallest subgroup of $G$ containing $X$)
I only have the answer for $G$ is finite group.
Edit: Sorry, now the question is right.
 A: The answer to your question is "no" in general. It is not enough in general.
Here's an example: we will work inside the group of invertible $2\times 2$ matrices with rational coefficients. Let $N$ be given by
$$N = \left\{\left.\left(\begin{array}{cc}
1 & m\\
0 & 1\end{array}\right)\in G\ \right|\ m\in\mathbb{Z}\right\},$$
and let
$$x = \left(\begin{array}{cc}
2 & 0\\
0 & 1\end{array}\right).$$
Let $G = \langle N,x\rangle$. Set $T=N\cup\{x\}$, and set $S=N$. Then trivially $N=\langle S\rangle$, and $G = \langle T\rangle$. Also, for every $n\in N$ we have $nNn^{-1}\subseteq n$ (you can verify that this is the case); and $xNx^{-1}\subseteq N$, since for every $m\in \mathbb{Z}$, we have:
$$\begin{align*}
\left(\begin{array}{cc}
2 & 0\\ 0 & 1\end{array}\right)\left(\begin{array}{cc}1 & m\\0 & 1\end{array}\right) \left(\begin{array}{cc}\textstyle\frac{1}{2}&0\\0&1\end{array}\right)
&= \left(\begin{array}{cc}
2 & 2m\\0 & 1\end{array}\right)\left(\begin{array}{cc}
\textstyle\frac{1}{2}&0\\0 & 1\end{array}\right)\\
&= \left(\begin{array}{cc}
1 & 2m\\
0 & 1\end{array}\right)\in N.
\end{align*}$$
However, $N$ is not normal, since
$$\begin{align*}
x^{-1}\left(\begin{array}{cc}1&1\\0&1\end{array}\right)x &= \left(\begin{array}{cc}
\textstyle\frac{1}{2} & 0\\0 &1\end{array}\right)\left(\begin{array}{cc}1 & 1\\ 0 & 1\end{array}\right)\left(\begin{array}{cc}2 & 0\\0&1\end{array}\right)\\
&= \left(\begin{array}{cc}
\textstyle\frac{1}{2}&\textstyle\frac{1}{2}\\0&1\end{array}\right)\left(\begin{array}{cc}2&0\\0&1
\end{array}\right)\\
&= \left(\begin{array}{cc}
1 & \textstyle\frac{1}{2}\\0&1\end{array}\right)\notin N.
\end{align*}$$
So we have $tSt^{-1}\subseteq N$ for all $t\in T$, but $N$ is not normal.
However, if $T$ is a generating set and $tNt^{-1}=N$ for all $t\in T$, then the conclusion holds. In particular, this gives your result in the case where $N$ is finite, since the hypothesis imply that $tNt^{-1}\subseteq N$ for all $t\in T$, hence (by finiteness of $N$), that $tNt^{-1} = N$, which in turn gives you $t^{-1}Nt=N$ as well.
The difficulty arises because in an infinite group, given a subgroup $N$, the subset
$$\{ g\in G\mid gNg^{-1}\subseteq N\}$$
need not be closed under inverses, as noted in this previous question from whence the example came.
