Is this an equivalent definition of a normal subgroup?

I believe there is an example of a non-normal subgroup satisfying (1) (and also (4)).

Let $G = \langle x,y \mid y^{-1}xy = x^2 \rangle$ and $N = \langle x \rangle$. Then $y^{-1}Ny < N$ but $yNy^{-1} \not< N$. (This group can also be defined as a group of $2 \times 2$ rational matrices.)

So, for all $k > 0$, $Ny^k < y^kN$ and $y^{-k}N < Ny^{-k}$ and hence $Ny^kN = y^kN$ and $Ny^{-k}N = Ny^{-k}$, which proves (1) and (4) when $g$ is a power of $y$.

The normal closure $K$ of $x$ in $G$ is an infinitely generated abelian group generated by elements $x_n = y^{-n}xy^n$ for $n \in \mathbb{Z}$, where $x_0=x$ and $x_n^2=x_{n+1}$. An arbitrary element $g \in G$ can be written as $y^kx_n^j$ for some $k,n,j$, or alternatively as $x_n^j y^k$ (with the same $k$ and $j$ but a different $n$). Since $K$ is abelian, each $x_n$ commutes with $N$, and so we get Condition (4) for all $g \in G$.

@Bartek, Geoff Robinson: Thanks for you remarks. I agree with them. My idea failed. But I have another and I corrected my text accordingly to it.:

It seems that you are right, and the condition (2) is equivalent to the normality of the group $N$. The necessity is obvious. Now suppose that the group $N$ satisfies condition (2). The set $NgN$ contains a left coset $gN$ and is equal to the left coset $xN$. Thus $NgN=gN=xN$. Then $g^{-1}NgN=g^{-1}gN=N$. This implies that $g^{-1}Ng\subset N$ for each $g\in G$, and, therefore the group $N$ is normal.