Find a normal matrix which commute with $A$. Consider
$$A=\begin{pmatrix}1&1&0&0\\-1&1&0&0\\0&0&2&2\\0&0&-2&2\end{pmatrix}.$$

I want to find a matrix $B$ such that

*

*$B$ is normal.


*$AB=BA$.


*$B\neq I$ and $B\neq \alpha A$ ( $\alpha \in \mathbb{R}^*$).

 A: $A$ has size 4 and 4 distinct eigenvalues.  It follows that a matrix $M$ will commute with $A$ if and only if $M = p(A)$ for some polynomial $p$.
In particular, every matrix which commutes with $A$ will have the form
$$
M = \pmatrix{a&b&0&0\\-b&a&0&0\\0&0&c&d\\0&0&-d&c}
$$
with $a,b,c,d \in \Bbb C$.
Notably, all matrices of this form are necessarily normal.
A: Since $A$ is normal, any scalar multiple of $A$ will do. If you want to look for the most general form, if you write $B$ in block form, 
$$
B=\begin{bmatrix} X&Y\\ Z&W\end{bmatrix} 
$$
and let $A_0=\begin{bmatrix} 1&1\\-1&1\end{bmatrix}$, then 
$$
AB=\begin{bmatrix} A_0&0\\0&2A_0\end{bmatrix}\begin{bmatrix} X&Y\\ Z&W\end{bmatrix} =\begin{bmatrix} A_0X&A_0Y\\ 2A_0Z&2A_0W\end{bmatrix} ,
$$
while
$$
BA=\begin{bmatrix} X&Y\\ Z&W\end{bmatrix}\begin{bmatrix} A_0&0\\0&2A_0\end{bmatrix} =\begin{bmatrix} XA_0&2YA_0\\ ZA_0&2WA_0\end{bmatrix}.
$$
We get $XA_0=A_0X$, and from this it is easy to check that $X=\begin{bmatrix} a&b\\-b&a\end{bmatrix}$. From $A_0Y=2YA_0$ and $ZA_0=2Z_0Z$, we get $Z=Y=0$, and from $2A_0S=2WA_0$, we get $W=\begin{bmatrix} c&d\\ -d&v\end{bmatrix}$. So 
$$
B=\begin{bmatrix} a&b&0&0\\-b&a&0&0\\ 0&0&c&d\\ 0&0&-d&c\end{bmatrix}
$$
for any choice of $a,b,c,d$. For any such choice, $B$ is already normal. 
