I redo Zev's exquisite answer for more space.
No, it does not mean that we always have $N\subseteq gNg^{-1}$ for any subgroup $N$ and any $g\in G$.
First, note that by the definition of set equality, 1) implies 2).
Why does 2) implies 1)? Note that $gNg^{-1}\subseteq N$ for some particular $g\in G$
$\iff$ Let $h=g^{-1}$ iff $h^{-1} = g$ $ \iff h^{-1}Nh\subseteq N$.
Thus, if for every $g\in G$, we have $\color{blue}{gNg^{-1}\subseteq N}$,
then also for every $g\in G$, we have $g^{-1}Ng\subseteq N$,
because every element of $G$ is some other element's inverse.
But by left and right multiplying by $g$ and $g^{-1}$, respectively, $g^{-1}Ng\subseteq N \iff N\subseteq gNg^{-1}$ . Therefore, 2) implies that for all $g\in G$, we have both $\color{blue}{gNg^{-1}\subseteq N}$ and $N\subseteq gNg^{-1}$ $\iff N=gNg^{-1}$.