Is this a counterexample for the Burnside's Theorem? The following text is trying to solve the number of distinct ways to color a square by two colors B and W :

The above text considers "the number of distinct ways to color a square by two colors B and W" as the number of distinct orbits. But according to the same book : 

If $G$ acts on a set $X$ and $x,y∈X$, then $x$ is said to be $G$-equivalent to $y$ if there exists a $g∈G$ such that $gx=y$. We write $x∼_Gy$ or $x∼y$ if two elements are $G$-equivalent.
If $X$ is a $G$-set, then each partition of $X$ associated with $G$-equivalence is called an orbit of $X$ under $G$. We will denote the orbit that contains an element $x$ of $X$ by $O_x$.
Example 6. Let $G$ be the permutation group defined by $G = {\{(1), (123), (132), (45), (123)(45), (132)(45)}\}$ and $X ={\{1,2,3,4,5}\}$. Then $X$ is a $G$-set. The orbits are $O_1 = O_2 = O_3 = 
{\{1,2,3}\}$ and $O_4 = O_5 ={\{4,5}\}$. 

My question is: In the case of coloring a square, elements of an orbit $O_x$ means the elements of $\widetilde X$ such that $gx=x$ or $x∼x$ because it includes the elements (maps) that doesn't changes coloring-arrangement. But original definition of orbit as well as a proof of the Burnside's Theorem in the same book says any $y\in \widetilde X$ such that $x∼y$ which in coloring example means "the maps such that if some rigid motion acts the result is another map" which means number of distinct orbits is $1$. Is this a counterexample for the Burnside's Theorem?  
A clear simple explanation would be much appreciated. 
PS - Source is the book Abstract Algebra by T. W. Judson. 
 A: The answer, that there are six ways to color, is correct, since the only possible colorings are
1) All vertices colored black.
2) All vertices colored white.
3) One vertex black, three white.
4) One vertex white, three black.
5) diagonally opposite two vertices black, other white.
6) Adjacent two vertices are black, other two white.
The problem you may be facing is understanding equivalent colorings or understanding orbit of a coloring. 
Label the vertices by $x,y,z,w$. Then in the coloring of vertices by black and white, what are possible ways?
$x$ can be B or W, $y$ can be $B$ or $W$, and so on. Thus, initially, we can do $2.2.2.2$ colorings.
We want to say, among them, which are different? Here, you are saying two colorings are same if (and only if) one can be obtained from other by rotation or reflection (i.e. by applying group element in the group of symmetries of square).
For example, following labelling (one black, other white) were different in beginning, but after taking group action, they are coming in same orbit, i.e. they are equivalent. 

With this understanding of same or different colorings, the answer, that there are six different colorings, is correct.
