Not all unitary matrices are orthogonal. 
Recall that a matrix $A\in \mathbb{C}^{n\times n}$ is normal if $AA^{*}=A^{*}A$ where $A^*:=\bar{A}^T.$ Let $A\in \mathbb{R}^{n\times n}.$



*

*Show that not all unitary matrices are orthogonal.

*Use 1. to conclude that not every normal matrix in $\mathbb{R}^{n\times n}$ is orthogonally similar to a diagonal matrix.


My idea for 1.: 

We want to show that $AA^T\neq A^TA \, \forall A=UBU^*$ where $U$ is a unitary matrix and $B$ is a diagonal matrix. $U^*=\bar{U}^T.$
  Then we have 
  \begin{align}
AA^T\\
&=UBU^*(UBU^*)^T\\
&=UBU^*\bar{U}BU^T
\end{align} So, intuitively since $U^*\bar{U}\neq \bar{U}U^*$, we have that $AA^T\neq A^TA .$ How do I come up with a clever counterexample? 

 A: Let $a, b \in \mathbb{C}\setminus\mathbb{R}$ such that
$$
a^2+b^2 \in \mathbb{C}\setminus\mathbb{R},\ |a|^2+|b|^2=1, \ a\bar{b} \in \mathbb{R}.
$$
Setting
$$
A=\left[\begin{array}{cc}a&b\cr -b&a\end{array}\right],
$$
we have
$$
AA^*=\left[\begin{array}{cc}a&b\cr -b&a\end{array}\right]\cdot\left[\begin{array}{cc}\bar{a}&-\bar{b}\cr \bar{b}&\bar{a}\end{array}\right]
=\left[\begin{array}{cc}1&0\cr 0&1\end{array}\right] ,
$$
but
$$
AA^T=\left[\begin{array}{cc}a&b\cr -b&a\end{array}\right]\cdot\left[\begin{array}{cc}a&-b\cr b&a\end{array}\right]
=\left[\begin{array}{cc}a^2+b^2&0\cr 0&a^2+b^2\end{array}\right]
\ne \left[\begin{array}{cc}1&0\cr 0&1\end{array}\right].
$$
A: Observe that a matrix $A$ is unitary iff $AA^*=A^*A=I$. Thus, every unitary matrix is normal. $A$ is an orthogonal matrix iff $AA^T=A^TA=I$ (every orthogonal matrix is also normal).
This way, if $A$ is an unitary and orthogonal matrix, then $A^*A=I=A^TA$, which means that $A^*=A^T$. But $A^*=\overline{A}^T$. Then, for an unitary orthogonal matrix $\overline{A^T}=A^T$. This means that all entries in $A^T$, and therefore in $A$, must be real. Just simply take an unitary matrix with a non real entry.
For example $i.B$ with $B$ orthogonal.
