Let $U$ be the upper unipotent subgroup of of $GL_n$. It is said that $$ U \cap wUw^{-1} = \{ (a_{ij}) \in U \mid a_{ij}=0, i<j, w^{-1}(i) > w^{-1}(j) \}. $$
How to prove this? I try to compute some examples in the case of $GL_3$. Let $$ u = \left(\begin{array}{ccc} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{array}\right) \in U. $$
Let $$ w= s_1 s_2 s_1 = \left(\begin{array}{ccc} 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0 \end{array}\right). $$ Then $$ wuw^{-1} = \left(\begin{array}{ccc} 1 & 0 & 0\\ c & 1 & 0\\ b & a & 1 \end{array}\right). $$ If we want $wuw^{-1} \in U$, then $a=b=c=0$. It seems that $U \cap wUw^{-1} = \{I\}$, where $I$ is the identity matrix. I don't know where am I wrong. Thank you very much.
You will get different answers, of different sizes, depending on which permutation $w$ you choose. Your computation so far is correct, by the way: $a_{12} = 0$ because $1 < 2$ and $w^{-1}(1) = 3 > 2 = w^{-1}(2)$, and so on.
To prove this in general, I'd suggest the following: remember that multiplying $u$ on the left by a permutation matrix will rearrange the rows of $u$, and multiplying on the right will rearrange the columns. From this you can (and should) figure out which entries $a_{ij}$ of $u$ (for $i<j$) must be zero in order for $w u w^{-1}$ to be upper-triangular. Where does the entry $a_{ij}$ end up? When does it land above the main diagonal?
(For example, in general conjugating by the permutation $w = (n,n-1,\ldots,2,1)$ will reverse the order of both the rows and the columns, i.e. it will transpose the matrix, which explains why $U \cap wUw^{-1} = \{I\}$ in that case. On the other extreme, if $w$ is the identity permutation, then obviously $U \cap wUw^{-1} = U$.)