Conjugates of the upper triangular matices It's a shame...I want to give an explicit description of the set $\bigcap_{m \in GL_n(K)} mUm^{-1}$, $U$ being the upper triangular subgroup of $GL_n(K)$. It seems to be $K^\times I_n$ but I do not know how to show the nontrivial inclusion.
 A: If $A$ is in the intersection, then for all $M\in GL_n(K)$ the matrix
$$
MAM^{-1}
$$
is upper triangular. With $M=I_n$, it follows that $A$ is upper triangular.
Take
$$
P = \begin{pmatrix}
0 & \dots & 0 & 1\\
0 & \dots & 1 & 0\\
\vdots&  && \vdots\\
1 & \dots & 0 & 0
\end{pmatrix},
$$
which implies $P^2=I_n$.
Then the mapping $A \mapsto PAP^{-1}$ reverses the orders of rows and columns.
Thus if $A$ is upper triangular, the matrix $PAP^{-1}$ is lower triangular.
It follows that $A$ must be diagonal: assume $a_{ij}\ne 0$ for some $i<j$.
Then the entry at position $(j,i)$ of $PAP^{-1}$ is non-zero, hence cannot be upper-triangular.
Now let assume that $A$ is not equal to an invertible multiple of the identity.
Then there are $i,j$ such that $a_{i,i} \ne a_{j,j}$.
Take matrix 
$$
Q=\begin{pmatrix} 1 & 0 \\ 1 &  1\end{pmatrix},
\quad Q^{-1}=\begin{pmatrix} 1 & 0 \\ -1 &  1\end{pmatrix},
$$
And it holds
$$\begin{split}
Q\begin{pmatrix} a_{ii}&0\\0&a_{jj}\end{pmatrix}Q^{-1}
&=\begin{pmatrix} 1 & 0 \\ 1 &  1\end{pmatrix}\begin{pmatrix} a_{ii}&0\\0&a_{jj}\end{pmatrix}\begin{pmatrix} 1 & 0 \\ -1 &  1\end{pmatrix}\\
&=\begin{pmatrix} 1 & 0 \\ 1 &  1\end{pmatrix}\begin{pmatrix} a_{ii}&*\\-a_{jj}&*\end{pmatrix}\\
& = \begin{pmatrix} *&*\\a_{ii}-a_{jj}&*\end{pmatrix},
\end{split}$$
which is not upper-triangular. Hence, one can construct a matrix $Q$ such that $QAQ^{-1}$ is not upper-triangular. It follows that $A$ is a multiple of the identity.
A: Two silly generalizations: 


*

*@daw's argument holds for any reductive algebraic group (take $U$ to be the Borel subgroup, take $P$ to be a preimage of the longest root, take $Q$ to be a generator from a root subgroup of the corresponding simple root).

*Assuming $U$ is upper triangular is pretty strong. An argument similar to @daw's actually only needs it to have some sort of block structure (replace Borel with parabolic). Let $U_i$ be the subgroup of block upper triangular matrices $\begin{bmatrix} A & B \\ 0 & D \end{bmatrix}$ with $A$ a $i \times i$ block. If we take $i=1$ then we get an easier linear algebra proof (that works for the original question):

Let $A \in U_1$. Then $A\vec{e}_1 = a_{11} \vec{e}_1$ so $\vec{e}_1$ is an eigenvector. If $A$ is in the intersection, then for every invertible matrix $M$, $M^{-1}AM$ has $\vec{e}_1$ as an eigenvector, so $M\vec{e}_1$ is an eigenvector of $A$ as well. Since eigenvectors for distinct eigenvalues are linearly independent, we must have that $A$ has a basis of eigenvectors, all of which have the same eigenvalue. In other words, $A$ is scalar.

If we don't mind a little representation theory, then we can do the same for any $U_i$. The argument now shows every $i$-dimensional subspace is invariant, but the intersection of invariant subspaces is invariant, so every $1$-dimensional subspace is invariant and $A$ is scalar.
Similar statements should hold over the integers (the analogues of $P$ and $Q$ are independent of the ring) and so over any ring. The linear algebra (eigenvetors and invariant subspaces) proofs won't go as smoothly, since not every vector is of the form $M\vec{e}_1$ for invertible $M$.
