invariant subspaces of orthogonal transformation I have an orthogonal transformation $\mathbb{A}: V=\begin{pmatrix}x\\ y\\z\end{pmatrix}\in\mathbb{R}^3\mapsto V'=\begin{pmatrix}
    0 & 0 & -1 \\
    1 & 0 & 0  \\
    0 & 1 & 0
  \end{pmatrix}\begin{pmatrix}x\\ y\\z\end{pmatrix}+\begin{pmatrix}1\\ 0\\0\end{pmatrix}\in\mathbb{R}^3$
Let A be the matrix of $\mathbb{A}$. It has eginevalues $\sqrt[3]{-1}$
If I want to find invariant subspaces(point, line or surface) I need to find an invariant of $\mathbb{A}$. That is solution of $\lambda$V=AV+B but it has no solution.
So how to find invariants?
If you need I can print an answer.
UPD: 
$\lambda=-1, \text{ then } -\begin{pmatrix}x\\ y\\z\end{pmatrix}=\begin{pmatrix}
    0 & 0 & -1 \\
    1 & 0 & 0  \\
    0 & 1 & 0
  \end{pmatrix}\begin{pmatrix}x\\ y\\z\end{pmatrix}+\begin{pmatrix}1\\ 0\\0\end{pmatrix},\text{ then } (A+E)V=\begin{pmatrix}-1\\ 0\\0\end{pmatrix}\\
\text{We have }\begin{pmatrix}
    1 & 0 & -1 \\
    1 & 1 & 0  \\
    0 & 1 & 1
  \end{pmatrix}\begin{pmatrix}x\\ y\\z\end{pmatrix}=\begin{pmatrix}-1\\ 0\\0\end{pmatrix}.\\\text{ Lets add third equation to first. We'll get:} \begin{pmatrix}
    1 & 1 & 0 \\
    1 & 1 & 0  \\
    0 & 1 & 1
  \end{pmatrix}\begin{pmatrix}x\\ y\\z\end{pmatrix}=\begin{pmatrix}-1\\ 0\\0\end{pmatrix}\\
\text{There is no solution if } x+y=-1 \text{ and } x+y=0$
 A: To begin with, your transformation is affine. Its linear part is orthogonal, but with the additional translation the overall transformation is not orthogonal. On the other hand, it is an isometry.  
Examining the linear part of $\mathbb A$, you’ve already found its eigenvalues to be $-1$, $-\omega$ and $-\omega^2$, where $\omega=\frac12(-1+i\sqrt3)$ is a primitive cube root of $1$. The complex conjugate pair indicates that there’s a rotation involved, while the $-1$ tells us that there’s also a reflection. The corresponding eigenvectors are $(1,-1,1)^T$, $(\omega^2,-\omega,1)^T$ and $(\omega,-\omega^2,1)^T$. You can find the axis of rotation by computing the orthogonal complement of the span of the two imaginary eigenvectors. This turns out to be the real vector $(1,-1,1)^T$, which happily coincides with the reflection direction. The rotation angle is just the argument of the complex eigenvalues (although getting the direction right requires a bit of additional work). So, we can characterize $\mathbb A$ as a rotation about $(1,-1,1)^T$ through an angle of $\frac{2\pi}3$ followed by a reflection in the plane perpendicular to this axis and finished off with a one-unit translation in the positive $x$-direction. The invariants of $\mathbb A$ are going to be a line parallel to $(1,-1,1)^T$ and a plane orthogonal to it. To find them, one can try finding a fixed point of $\mathbb A$, i.e., solving $\mathbb A\mathbf v=\mathbf v$. If it exists, this fixed point lies on both the invariant line and plane, so once that’s been found, the invariant line and plane are easily constructed.  
The equation $\mathbb A\mathbf v=\mathbf v$ expands into the system $$\begin{align} x+z&=1 \\ x-y&=0 \\ y-z&=0\end{align}$$ with solution $\left(\frac12,\frac12,\frac12\right)$. The invariant plane is therefore $2(x-y+z)=1$ and the line is $\left(\frac12,\frac12,\frac12\right)^T+t\,(1,-1,1)^T$.  
I would go about this a different way, however. Since $\mathbb A$ is affine, it seems natural to move from $\mathbb R^3$ to the projective space $\mathbb P^3$ and work in homogeneous coordinates. $\mathbb A$ becomes a linear transformation and its invariants are its eigenspaces and their sums. Determining them is a largely a matter of applying the familiar operations of computing bases for spans and null spaces.  
The homogeneous matrix for $\mathbb A$ is $$\begin{bmatrix}0&0&-1&1\\1&0&0&0\\0&1&0&0\\0&0&0&1\end{bmatrix}$$ with eigenvalues $1$, $-1$, $-\omega$, $-\omega^2$ and eigenvectors $\mathbf v_{1} = (1,1,1,2)^T$, $\mathbf v_{-1} = (1,-1,1,0)^T$, $\mathbf v_{-\omega} = (\omega^2,-\omega,1,0)^T$ and $\mathbf v_{-\omega^2} = (\omega,-\omega^2,1,0)^T$, respectively.  
These eigenvectors and their associated eigenspaces are the fixed points of $\mathbb A$. Only $\mathbf v_1$ is a finite point, $\left(\frac12,\frac12,\frac12\right)^T$ in inhomogeneous Cartesian coordinates. $\mathbf v_{-1}$ is a point at infinity that corresponds to the direction of the reflection, while the other two are imaginary points at infinity related to the rotation.  
Taking pairwise sums, we get the invariant lines of $\mathbb A$. The only ones that produce real lines are $\lambda \mathbf v_1+\mu \mathbf v_{-1}$—the line $\left(\frac12,\frac12,\frac12\right)^T+t\,(1,-1,1)^T$—and $\lambda \mathbf v_{-\omega}+\mu \mathbf v_{-\omega^2}$, which corresponds to the real line at infinity that is is intersection of the planes $(1,-1,1,\tau)^T$ with the plane at infinity, i.e., the intersection of all of the planes that are orthogonal to the rotation axis. Computing a basis for the row space of $\begin{bmatrix}\mathbf v_{-\omega}&\mathbf v_{-\omega^2}\end{bmatrix}^T$ via row-reduction yields a different representation of this line as the span of $(1,0,-1,0)^T$ and $(0,1,1,0)^T$, which makes it more obvious that this is indeed a real line, albeit one at infinity.   
Moving to invariant planes, we take eigenvectors three at a time. Again, there are only two combinations that result in real planes: $\lambda\mathbf v_1+\mu\mathbf v_{-\omega}+\tau\mathbf v_{-\omega^2}$ produces the plane at infinity $(0,0,0,1)^T$, which is an invariant space of every affine transformation; and $\lambda\mathbf v_{-1}+\mu\mathbf v_{-\omega}+\tau\mathbf v_{-\omega^2}$, which is the plane $(2,-2,2,-1)^T$, i.e., $2(x-y+z)=1$, as found above. That these sums produce real planes isn’t all that surprising, since we’ve found previously that the span of $\mathbf v_{-\omega}$ and $\mathbf v_{-\omega^2}$ is a real line.
