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Let $P_1$ be a irregular polygon with $n$ sides, $P_1$ is given by

\begin{equation} P_1 = \begin{bmatrix} x_1 \quad x_2 \quad \ldots \quad x_n \\ y_1 \quad y_2 \quad \ldots \quad y_n \end{bmatrix}. \end{equation}

Let $C_x$ and $C_y$ be center of $P_1$ in $x$ and $y$ axes, respectively. And let $P^{*}_1$ be a regular polygon constructed by $P_1$, given by

\begin{equation} P_1^{*} = \left(C_x + r\times \cos\left( \frac{2\pi i}{n}\right), C_y + r\times \sin\left( \frac{2\pi i}{n}\right)\right), \end{equation} for $i = 1, 2, \ldots, n$ and $r = 1$ (radius).

How to transform $P_1^{*}$ in $P_1$? For example, $P_1 = T \times P_1^{*}$, where $T$ is a transformation matrix.

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    Do you really mean to say that $r=1$? If so, why not just write $P_1^*=\left(C_x + \cos\left(\frac{2\pi i}{n}\right), C_y + \sin\left(\frac{2\pi i}{n}\right)\right)$? – 2017-01-17
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    Because this information would be important. – 2017-01-17
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    All the same information is there if you do not write $r$ at all. It would describe the exact same polygon. But if you want to especially _alert_ people to the fact that the polygon has radius $1$, you could write, "let $P_1^*$ be a regular polygon of circumradius $1$ ... ." – 2017-01-17

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$$ \newcommand{\bx} {\mathbf x} \newcommand{\by} {\mathbf y} \newcommand{\bz} {\mathbf z} \newcommand{\bw} {\mathbf w} \newcommand{\bu} {\mathbf u} \newcommand{\bv} {\mathbf v} $$ Let $\bx$ denote the column vector with entries $x_1, \ldots, x_n$, and similarly for $\by$, where these are the coordinates of the irregular polygon. Let $\bz$ and $\bw$ denote column vectors filled with the $x$ and $y$ coordinates of the regular polygon.

Now let $$ \bu = \frac{1}{s} \left(\bx - \frac{\bx \cdot \by}{\by \cdot \by} \by \right);\\ \bv = \frac{1}{t} \left(\by - \frac{\bx \cdot \by}{\bx \cdot \bx} \bx \right), $$ where $$ s = (\bx \cdot \bx) - \frac{(\bx \cdot \by)^2}{\by \cdot \by}\\ t = (\by \cdot \by) - \frac{(\bx \cdot \by)^2}{\bx \cdot \bx}. $$

Note, through direct computation, that $\bu\cdot \bx = 1; \bu \cdot \by = 0; \bv \cdot \bx = 0; \bv \cdot \by = 1$.

Let $M$ be the $2 \times n$ matrix with rows $\bu$ and $\bv$. And let $K$ be the $n \times 2$ matrix with columns $\bz, \bw$. Then $$ KM $$ takes the vector $\bx$ to the vector $\bz$, and $\by$ to $\bw$. But you wanted it to multiply on the right, with row vectors, so the matrix $T$ you seek is $$ M^t K^t $$ Or at least, that's ONE possible matrix that does what you seek.

Note that it's essential that the vectors $\bx$ and $\by$ not be parallel for this approach to work. (Indeed, if they are parallel, there's no solution.)

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    I sorry, I can't answer, because some problems. I work in a example given by: Let $P_1$ a irregular polygon and $\mathbf{x} = (1, 2, 3, 4)$ and $\mathbf{y} = (1,3,4,2)$ yours coordinates. Let $P_2$ a regular polygon given by $\mathbf{z} = (2.58, 1.58, 2.58, 3.58)$ $\mathbf{w} = (3.38, 2.38, 1.38, 2.38)$. The inner product $\mathbf{u} \cdot \mathbf{x} = 1.9 \neq 0$. I'm thinking about the answer. – 2017-01-19
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    Apologies; I've fixed the expressions for $s$ and $t$, which had mistakes in them. – 2017-01-19
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First, $T$ will have to go on the left, not the right. Second, you need to express $P_1$ in homogeneous coordinates, because otherwise the translation part (i.e., adding $(C_x, C_y)$ to each point) will not be representable by a linear transformation. So rewriting:

\begin{equation} P_1 = \begin{bmatrix} x_1 & x_2 & \ldots & x_n \\ y_1 & y_2 & \ldots & y_n \\ 1 & 1 & \ldots & 1 \end{bmatrix} \end{equation} you can write

\begin{equation} Q = \begin{bmatrix} c & -s & C_x \\ s & c & C_y \\ 0 & 0 & 1 \end{bmatrix}\cdot A\cdot B\cdot C \end{equation} where $c = \cos (\frac{2\pi}{n}); s = \sin(\frac{2\pi}{n})$, and $A, B,$ and $C$ are the matrices \begin{equation} A = \begin{bmatrix} u & -v & 0 \\ v & u & 0 \\ 0 & 0 & 1 \end{bmatrix}; B = \begin{bmatrix} \frac{1}{s} & 0 & 0 \\ 0 & \frac{1}{s} & 0\\ 0 & 0 & 1 \end{bmatrix}; C = \begin{bmatrix} 1 & 0 & -C_x \\ 0 & 1 & -C_y \\ 0 & 0 & 1 \end{bmatrix}. \end{equation} Here $s = \sqrt{(x_1 - C_x)^2 + (y_1 - C_y)^2)}$.

(BTW, if you decide you want a radius other than $1$, use $r/s$ instead of $1/s$ in the matrix $B$).

Now all I have to tell you is what $u$ and $v$ are. They are the cosine and sine (respectively) of $-\theta$, where $$ \theta = atan2(y_1 - C_y, x_1 - C_x). $$

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    It may be worth mentioning that the reason for the $c$ and $s$ entries in the matrix is that the question defined $P_1^*$ in such a way that it assigned the coordinates $(C_x+1, C_y)$ to its last vertex rather than to its first vertex. To make those the coordinates of the first vertex, replace $i$ with $(i-1)$. Another alternative is to make a single rotation by $\frac{2\pi}{n} - \theta$ rather than one by $-\theta$ and one by $\frac{2\pi}{n}$. – 2017-01-17
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    @John Is $Q$ the transformation matrix? If yes, then, we have a problem. The dimension of $Q$ is $3 \times 3$ and $P_1{*}$ is $3 \times n$. – 2017-01-17
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    Wagner: $QABC$ is the matrix, and yes, it has to go on the left, i..e, $P_1^{*} = T P_1$. (I see that this is actually the reverse of what you wanted, but since each of the factors is easy to invert, you should be able to derive the thing you want.) – 2017-01-17
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    Actually, DavidK, I'd failed to note that $i$ goes from $1$ to $n$ rather than $0$ to $n-1$....but it turns out you're right. :) And yes, they could be combined, but it seemed as if there was some hope of having OP understand the derivation if I kept them separate. – 2017-01-17
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    I think you are right that two (simple) rotations are a better explanation than one (harder to explain) rotation. – 2017-01-17
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    Wagner: it's possible you really DO want $T$ on the right, but that is such a weird question that I'm not going to try to answer it, esp. since the space of answers has a very large dimension (roughly $(n-2)^2$), while the correct answer for your question (as I've interpreted it) is unique. – 2017-01-17
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    I didn't explain correctly, but answer is correctly. Sorry, for I don't have a mathematical language. :) Thank you so much for your answer. Thank you! – 2017-01-17
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    I understood because the question caused confusion. I corrected the first line of question. $P_1$ is a \textbf{irregular} polygon. – 2017-01-17