Eigenvectors for a sum of diagonal and anti-diagonal matrices Consider the case of invertible matrix which is the sum of diagonal and anti-diagonal matrices, e.g.,
$$\begin{bmatrix} \color{red}1 &   0 &  0 &  0 &  \color{red}6 \\ 0 &  \color{red}2 &  0 &  \color{red}7 &  0 \\ 0  & 0 & \color{red}3 &  0 &  0 \\ 0 &  \color{red}8  & 0 & \color{red}4 &  0 \\ \color{red}9 &  0  & 0 & 0 &  \color{red}5\end{bmatrix}$$
Such matrices I name shortly $X$-matrices (even shorter $X$ - do they have more official name?) and it's easy to check that the sum of two $X$-matrices is an $X$-matrix.
Also the product of two matrices is   an $X$-matrix as
$$X_1X_2=(D_1+A_1)   (D_2+A_2)=(D_1D_2+A_1A_2)+(D_1A_2+A_1D_2)$$
($D,A$ denoted here as diagonal and antidiagonal part of $X$) and product of two diagonal or two anti-diagonal is always diagonal and product of diagonal and anti-diagonal is anti-diagonal.
Further if $X$-matrix is invertible also its inverse is an $X$-matrix because inverse can be presented as a polynomial of $X$ from Cayley-Hamilton theorem.
Making calculations I have found one more property of these matrices:
i.e. also eigenvectors $v_1, v_2, \dots $ for this type of matrix can be grouped  to make $X$-matrix.
For instance for the matrix listed above we have eigenvectors as colummns of
$$V=\begin{bmatrix}
 \color{red}{-0.730}  &  \color{red}{0.529}   &  0.000  &   0.000   &  0.000 \\
  0.000  &  0.000   &  \color{red}{0.730}  &  \color{red}{-0.633}  &   0.000  \\
  0.000 &   0.000   & 0.000  &   0.000  &  \color{red}{1.000}  \\
  0.000  &   0.000  & \color{red}{-0.683}  &  \color{red}{-0.774}  &  0.000  \\
  \color{red}{0.683} &   \color{red}{0.848}   &  0.000  &   0.000  &   0.000  \\
\end{bmatrix}$$
and it's possible to permute columns in order to obtain  from them an $X$-matrix.

*

*How this last property can be proved? How  can we prove that there is a permutation of eigenvectors of $X$-matrix which is also an $X$-matrix?

*Could we use for proof the equation $X=VDV^{-1}$ where however $V$, if columns are chosen randomly, can be in the form which is not an $X$-matrix? (but its some permutation supposedly is ...)

 A: Your X-matrices are actually a pretty well known family in disguise: block diagonal matrices with blocks $2\times2$. Go read that paragraph in Wikipedia, it is pretty short, and whatever you need from it is shorter yet:

The eigenvalues and eigenvectors of ${A}$ are simply those of $A_{1}$ and $A_{2}$ and ... and $A_{n}$ (combined).

Indeed, in an X-matrix, $x_1$ interacts only with $x_n$, $x_2$ with $x_{n-1}$, and so on. Why wouldn't you reorder your basis vectors so as to put the interacting ones next to each other? With your $5\times5$ example, this implies reordering the basis as $(x_1,x_5,x_2,x_4,x_3)$. The matrix which does that is $$P=\begin{pmatrix}1& 0& 0& 0& 0\\
0& 0& 0& 0& 1\\
0& 1& 0& 0& 0\\
0& 0& 0& 1& 0\\
0& 0& 1& 0& 0\end{pmatrix}$$
Now apply that to your matrix $X$ and get
$$PXP^T=\begin{pmatrix}1& 6& 0& 0& 0\\
9& 5& 0& 0& 0\\
0& 0& 2& 7& 0\\
0& 0& 8& 4& 0\\
0& 0& 0& 0& 3
\end{pmatrix}$$
Q.e.d.
A: Hint. In your 5x5 matrix, you can easily observe that there ALWAYS exist:
a. Two eigenvectors of the form $u=(x,0,0,0,y)$ - because $Au=(x',0,0,0,y')$
b. Two eigenvectors of the form $v=(0,x,0,y,0)$ - because $Av=(0,x',0,y',0)$ and
c. the eigenvector $(0,0,1,0,0)$.
Put them together as columns and obtain an X-matrix.
