Simply note that $SO(n)$ is the kernel of the group homomorphism
$$ \det :O(n) \longrightarrow \mathbb{R}^* \ \ \ \ \ A \mapsto \det A$$
Recall that every kernel is a normal subgroup.
As mentioned, the definition of normal subgroup specifies that
$g Ng^{-1} = N \tag 1$
for all $g \in G$. We note this entails
$\forall g \in G, \; g^{-1}N g = g^{-1}N(g^{-1})^{-1} = N; \tag 2$
therefore, though the mappings
$g \mapsto gng^{-1} \tag 3$
and
$g \mapsto g^{-1}ng \tag 4$
are not in general the same, they give rise to the same definition of normal subgroup, since conjugation by every $g \in G$ is required; thus we always pick up both left and right conjugation by any paricular $g$. So in fact the choice of right or left conjugation is arbitrary when defining normal subgroup.
In closing, note that left and right conjugation are the same, that is,
$gng^{-1} = g^{-1}ng \tag 5$
if and only if
$g^2n = ng^2, \; \forall n \in N, \; g \in G; \tag 6$
i.e., when ever every element of $N$ commutes with every square in $G$, when the set of squares is contained in $C_N(G)$, the centralizer of $N$ in $G$.
Best Answer
Let's take a look at the group of rotations of cube. It has a subgroup of rotations around vertical axis. This subgroup (let's call it $A1$) has 4 elements: rotate the cube for 0, 90, 180 or 270 degrees.
There is another subgroup: rotation around one of horizontal axes. Let's call it $A2$.
Subgroups $A1$ and $A2$ are obviously different. But still they look so very much alike! If there was someone else looking at our cube from different angle he could even fail to understand my descriptions of $A1$ and $A2$ "correctly" and confuse $A1$ with $A2$.
This is because $A1$ and $A2$ are conjugated. The $g x g^{−1}$ actually means "look at $x$ from another point of view", and $g$ defines this "point of view".
Subgroup is normal if it is very "symmetric". No matter from which point you look at the whole group $G$, the subgroup $N$ remains at place.
UPDATE: example of a normal subgroup.
Let's take the same cube. Now lets allow only rotations for 180 degrees around $x$, $y$ and $z$ axis. And any combination of such rotations. This will be group $B$. $B$ is a subgroup of the group of all rotations of cube. It is a proper subgroup of the original group (each face of our cube either remains in place or is moved to the opposite position by our rotations, so not all the elements of original group are included into this subgroup).
The definition of $B$ "does not depend of frame of reference", so I am sure this is a normal subgroup. Well, I understand that "does not depend of frame of reference" is not an accurate description, but this whole question is about intuition.
By the way, looks like group $B$ consists of only 4 elements: after any combination of described rotations only 1 of 2 faces can become the front, and only 1 of 2 faces can become the top. So "and any combination of such rotations" in my definition of group $B$ can be substituted with "and identity element $e$".