$\scriptsize\text{Though these incorrect contradictory statements aren't well received, I just want it explained.}$
Given graph $G=K_6$ where all edges are red, pick three points and change the color of the three edges connecting them (red to blue, blue to red). Repeat this for an arbitrary number of times.
We count how many possible graphs we could get during this process. Call it $t$.
Since there're $\text C_6^2=15$ sides, each of which has two possible colors, we get $t\le2^{15}$.
On the other hand, if a triple of points were operated twice, the effect cancels, so WLOG each triple got operated $1$ or $0$ times. Collect all triples in a set $S$. Operating a subset of $S$ mean operating on all its elements. If the operation on two different subsets of $S$, $S_1$ and $S_2$, result in the same graph $G^\prime$, then operating on $S_1-S_2$ and $S_2-S_1$ will also give the same result. Therefore, operating on $S_1\bigtriangleup S_2$ leaves the graph unchanged. This means that $K_6$ can be separated into some $K_3$s, thats clearly impossible, contradiction.
This means that operating on each subset of $S$ gives a different graph, so $t\ge2^{|S|}=2^{20}$.
So we arrive at $2^{20}\le t\le2^{15}$, how am I wrong?
$\def\FF{\mathbb{F}}$Let $\FF_2$ be a field with two elements, and view the vector space $V=\FF_2^{15}$ as having the coordinates of its vectors indexed by the edges of the graph.
Each triple is a vector $t$ in $V$ with exactly three non-zero coordinates. There are $20$ of them — let us write them $t_1,\dots,t_n$. Since $V$ has dimension $15$ and there are $20$ triples, there are at least $5$ independent linear relations between them, and therefore $2^5$ linear relations in all. If $t_{i_1}+\cdots+t_{i-r}=0$ is one of them, then switching the colors of the edges according to the corresponding triples to the summands does not do anything.
This tells you that $2^{20}$ overcounts the number of graphs by a factor of at at least $2^5$.