No, I don't think it's possible. Start from any side of the $99$-gon and go around it in a cunterclockwise direction, until you reach the initial side again. At each passage from a side to a consecutive one, record the color change (for example, red-blue, which will be encoded by $rb$, or yellow-red, which we'll call $yr$, and so on). If $x,y$ represent colors ($r,b$ or $y$), we agree that $xy$ and $yx$ cancel each other. After this, we compute the sum of all these symbols (a symbol is made up of two letters, just as explained above). It's easy to see now that by performing the allowed transformations, the sum of the symbols remains invariant, because we change $rb,br$ into $ry,yr$, or $by,yb$ into $br,rb$, and so on. Two pairs which sum up to $0$ are always replaced by two pairs which also sum up to $0$. However, the sum of the symbols over the first polygon is $rb+by+yr$, while the sum over the second polygon is $br+ry+yb$, which is different from the first one. P.S. To formalize this, we considered the free abelian group on the three symbols $rb,by,yr\ (*)$, and we took $br,ry,yb$ to be the inverses of $(*)$ (in that order). Then, to each polygon we associated an element of this group which is invariant under the permitted transformations, and found that the two given polygons have different invariants.

I will assume both listings of colors in the problem are in clockwise order (can I do this?) Put $ 0$ on a red side, $ 1$ on a blue side, $ 2$ on a yellow side. The total number of increases from one side to another, going clockwise, remains unchanged after changing any side, and same with the total number of decreases. This is because any side that is changed must be sandwiched between two sides of the same color. The starting configuration has $ 50$ increases and $ 49$ decreases, and the ending one has $ 49$ increases and $ 50$ decreases, so it is impossible.

Here's my silly invariant: for the $i^{th}$ side ($1\le i\le99$), we label $a_i=0$ if the side is red, $a_i=1$ if the side is blue, and $a_i=2$ if the side is yellow. We can easily see that \[P=(a_2-a_1)(a_3-a_2)\cdots(a_{99}-a_{98})(a_1-a_{99})\pmod3\]does not change since all modifications turn three consecutive sides of colors $x,y,x$ into $x,z,x$ where $x\ne y,z$, but \[(y-x)(x-y)\equiv(z-x)(x-z)\equiv-1\pmod3\]since the only nonzero quadratic residue modulo $3$ is $1$. But now we see that the original $P$ is (starting with red and ending with yellow, $a_{2k-1}=0$ and $a_{2k}=1$, except for $a_{99}=2$), \[[(1)(-1)\cdots(1)(-1)(1)](1)(-2)\equiv(-1)^{\frac{97-1}{2}}(-2)\equiv1\pmod3\]while the new $P$ is (starting with blue and ending with yellow, $a_{2k-1}=1$ and $a_{2k}=0$, except for $a_{99}=2$), \[[(-1)(1)\cdots(-1)(1)(-1)](-1)(2)\equiv(-1)^{\frac{97+1}{2}}(-2)\equiv2\pmod3,\]which is a contradiction.

hello we overkill with group theory Start with the first side and go clockwise, recording each color change as $a$ for red to blue, $b$ for blue to red, $c$ for red to yellow, $d$ for yellow to red, $e$ for yellow to blue, and $f$ for blue to yellow, and represent a coloring as the product. The transformation rules are $ab = cd$, $ba = fe$, and $dc = ef$. Note that the elements $a = (123)$, $b = (34)$, $c = (23)$, $d = (12)(34)$, $e = (12)$, $f = (234)$ of $S_4$ satisfy these. The first coloring is $(ab)^{48}afd$ and the second is $(ab)^{48}ceb$, however $afd$ has even parity and $ceb$ has odd parity, so they cannot be the same element.

Hi, I'm very interesting about the last reply. I don't understand how you translate the initial constraints with your equalities (why $ab = cd$ ? and so on). Moreover, you say $a$ for red to blue. But at the beginning, what colour do you assume is colored the necklace ? Is it plain ? If so, then you can't go from "red to blue". Moreover, it seems very strange to me that $a=(123)$ et $b = (3 4)$ as, to me, they are invert of each other. Could someone clarify ?

No one for this ?

Label red, blue, yellow as $1, 0, -1$ respectively. So, from $1 \to 99$, we have a chain of\[\underbrace{1, 0, 1, 0, \ldots , 1, 0}_{98 \text{ vertices}}, -1\]say in the clockwise direction. We compute the difference of each side in the clockwise direction; that is, we have differences of\[\underbrace{1, -1, 1, -1, \ldots , 1, -1}_{96 \text{ sides}}, 1, 1, -2\]which have signs of\[\underbrace{+, -, +, -, \ldots , +, -}_{96 \text{ signs}}, +, +, -.\]Note that in general, upon a move, the number of each sign does not change; no moves are ever made on a vertex who is surrounded by two different colors. Indeed, we check that\[1, 0, 1 \to 1, -1, 1 \iff +, - \to +, -\]\[0, 1, 0 \to 0, -1, 0 \iff -, + \to +, -\]\[-1, 1, -1 \to -1, 0, -1 \iff -, + \to -, +\]and vice versa. Our original color configuration, when number-coded, has $50$ sides whose clockwise difference is positive. The final, expected color configuration, when number coded, is\[\underbrace{1, 0, 1, 0, \ldots , 1, 0}_{96 \text{ vertices}}, 1, -1, 0\]and has clockwise side differences of\[\underbrace{1, -1, 1, -1, \ldots , 1, -1}_{96 \text{ sides}}, 2, -1, -1\]of which there are $49$ positive numbers. $50 \neq 49$ so the final configuration is not attainable. $\blacksquare$

Does this work? Label the colors as 0,1,2, s.t. we want to transform (0,1,0,1,...,0,1,2) to (0,1,0,1,...,0,2,1). Consider the difference sequence $d_i=a_i-a_{i-1}$, and consider the signs of those; in the original sequence, there's 48 negatives, but in the new, we want 49 negatives. However, note that changing a number from 2,1,2 <-> 2,0,2 doesn't change the \# of negatives in the sequence, and analogously for 1,0,1<->1,2,1 and 0,1,0<->0,2,0. We're done.

The answer is no. Label the vertices $1$ through $99$, and let $x_i = 0$ if vertex $i$ is colored blue, $x_i = 1$ if $i$ is red, and $x_i = 2$ if $i$ is yellow. Let $y_i = x_i - x_{i-1}$ (where indices are cyclic) and consider the sequence $\{y_i\}$. Notice that upon performing the operation, two consecutive terms $(y_i, y_{i+1}) = (1, 2)$ are changed into $(y_i, y_{i+1}) = (2, 1)$ or vice versa. Thus the number of $1$'s in $\{y_i\}$ remains invariant. However, we wish to transform $$(2, 1, 2, 1, \dots, 2, 1, 2, 2, 2) \to (2, 1, 2, 1, \dots, 2, 1, 1, 1, 1)$$which in this case is obviously impossible.