Rotation in multidimensional space I have a regular pentagon, with coordinates 
\begin{align*}
    x(1) &= (2, 0, 1, 1, 0) \\
    x(2) &= (1, 1, 0, 2, 0) \\
    x(3) &= (0, 2, 0, 1, 1) \\
    x(4) &= (0, 1, 1, 0, 2) \\
    x(5) &= (1, 0, 2, 0, 1)
\end{align*}
the side of pentagon is 2. I would like to know if there is way to find hyperplane, so that this pentagon in respect to the hyperplane be 2 dimensional? 
Remark: the central point can be found by directive ways 
$$    \left(\frac{4}{5}+\frac{1}{5} \sqrt{5 \left(\frac{1}{\sin \left(\frac{\pi }{5}\right)}\right)^2-14}\right) (1,1,1,1,1)$$
 A: The figure you have described is not a regular pentagon.
A regular pentagon with side $2$ would have diagonal $1 + \sqrt 5$,
but the corresponding "diagonal" of this figure 
(for example, the segment from $x(1)$ to $x(3)$)
has length $\sqrt{10}$.
We can take five vectors parallel to the five sides of the "pentagon"
by taking the differences of coordinates of successive vertices,
$x(k+1) - x(k)$.  If the five vertices all lay in one two-dimensional plane
then these five vectors would all lie in a two-dimensional subspace
parallel to that plane.
But in fact we can take linear combinations of these five vectors
to form a set of four vectors such as the following:
$$
v_1 = \pmatrix{1 \\ -1 \\ 0 \\ 0 \\ 0}, \quad
v_2 = \pmatrix{0 \\ 1 \\ -1 \\ 0 \\ 0}, \quad
v_3 = \pmatrix{0 \\ 0 \\ 1 \\ -1 \\ 0}, \quad
v_4 = \pmatrix{0 \\ 0 \\ 0 \\ 1 \\ -1}, \quad
$$
These vectors span a four-dimensional subspace, so the five vertices do
not lie in a single plane.
So the five points do not form any object we would normally call a
pentagon, and the object they do form is four-dimensional.
You may want to clarify what you mean by "two-dimensional" in the question.
A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\vx}{\mathbf{x}}\newcommand{\vv}{\mathbf{v}}$Generally, if $\vx_{0}$, ..., $\vx_{k}$ are coplanar points of $\Reals^{n}$ such that $\vx_{0}$, $\vx_{1}$, and $\vx_{2}$ are non-collinear, then the displacement vectors $\vv_{1} = \vx_{1} - \vx_{0}$ and $\vv_{2} = \vx_{2} - \vx_{0}$ are linearly independent, and parallel to the plane containing all the points; that is, each $\vx_{j}$ is uniquely of the form
$$
\vx_{0} + s \vv_{1} + t \vv_{2}
\tag{1}
$$
for some scalars $s$, $t$.
The set of points of the form (1) is a parametric description of the plane you seek. (Since the plane lies in $\Reals^{5}$, you would need three equations to describe this plane as the solution set of a linear system.)
Presumably you'll take $\vx_{0}$ to be the center of the pentagon, and $\vx_{1}$, ..., $\vx_{5}$ to be the vertices. In this case, if you perform the Gram-Schmidt algorithm on the ordered basis $(\vv_{1}, \vv_{2})$, obtaining an orthonormal basis $(\Basis_{1}, \Basis_{2})$ with $\vv_{1} = r\Basis_{1}$, then for each $j = 1$, ..., $5$,
$$
\vx_{j} = \vx_{0} + (r\cos \theta_{j}) \Basis_{1} + (r\sin \theta_{j}) \Basis_{2},\qquad \theta_{j} = \tfrac{2(j-1)\pi}{5}.
$$
