I came across the following while doing some related proof;
It seems easy to prove. $\quad$  

We are in ${\mathbb{M}}\_n(\mathbb{C})$, $n>1$: 


$1$) Given a unitary $n\times n$ matrix $U$, there is some permutation of the columns such that the modulus of each entry on the diagonal of the resulting matrix is $\le \dfrac{\sqrt{2}}{2}$, $\quad$ 


And more difficult:


$2$) There are no $n\times n$ unitary matrices $U$ of diagonal $D$ with all entries satisfying $|d\_{i,i}|>\dfrac{1}{\sqrt{n-1}}$ for all $i$, unless $U$ is a direct sum of unitaries $U\_k$, (up to a permutation congruence) $U\_k\in {\mathbb{M}}\_k$, $k<n$. 
I am searching for a proof or related facts.  

Thanks.


Edit: for $2$ and $n=4$, we can take $\begin{pmatrix}a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a\end{pmatrix}$. 
The only thing is that it may(not) hold for $n=3$.



Unitary condition
Consider a unitary matrix $U = (u\_{i,j})$. We will show that there is some permutation $\pi:[n] \to [n]$ such that $|u\_{i,\pi(i)}| \leq \sqrt{2}/\sqrt{n+1}$ for all $i$, which is sufficient to prove the first stated conjecture when $n \geq 3$, and when $n=2$ we can use the expression of $U$ as:
$$\left(\begin{array}{cc}
\sin(\theta) & \cos(\theta)\\
\cos(\theta) & -\sin(\theta)
\end{array}\right)$$
to obtain the desired bound, since $\min(\sin(\theta),\cos(\theta)) \leq \sqrt{2}/2$ for any $\theta$. Note that the process of permuting columns doesn't preserve the trace of the matrix, but it does preserve the property of being unitary!


Given $\pi$, consider the associated quantity $I(\pi) = \sum\_{i=1}^n |u\_{i,\pi(i)}|^2$. Since there are only finitely many permutations, there is some permutation $\pi\_\*$ minimizing this value. In particular, letting $v\_{i,j} = u\_{i,\pi\_\*(j)}$, the resulting matrix $V= (v\_{i,j})$ after applying the permutation of the columns must satisfy
$$|v\_{i,j}|^2 + |v\_{j,i}|^2 \geq |v\_{i,i}|^2 + |v\_{j,j}|^2$$
for all indices $i,j$, since otherwise we could exchange rows $i$ and $j$ and obtain a smaller value of $I(\pi)$.


Consider summing the above inequality over $j$. On the left hand side we get the squared norm of a row and a column of $V$, which both must be $1$. Thus we get:
$$2 \geq n|v\_{i,i}|^2 + I(\pi\_\*).$$
Since $I(\pi\_\*) \geq |v\_{i,i}|^2$, this gives the claimed bound.


Summing the latter inequality over $i$, we moreover deduce that:
$$2n \geq 2nI(\pi\_\*),$$
whence we conclude that $I(\pi\_\*) \leq 1$. Thus in particular, we **cannot** have $|v\_{i,i}| > 1/\sqrt{n}$ for every $i$, which resolves one possible interpretation of the second part of the OP's question.


It is worth noting that it might be possible to strengthen these bounds for large enough $n$ by considering a more interesting permutation than a transposition in the original inequality, but it is tight for $n=2$ by the above and for $n=3$ since there is no way to avoid having a $\sqrt{2}/2$ on the diagonal of a matrix whose columns are a permutation of:
$$\left(\begin{array}{ccc}
0 & \sqrt{2}/2 & \sqrt{2}/2\\
0 & \sqrt{2}/2 & -\sqrt{2}/2\\
1 & 0 & 0
\end{array}\right)$$


