I would like to take a general 5 x 5 matrix which is a triangular matrix "upside down", i.e.
$$A = \begin{pmatrix} 0 & 0 & 0 & 0 & a\\ 0 & 0 & 0 & b & c\\ 0 & 0 & d & e & f \\ 0 & g & h & i & j \\ m & n & o & p & q \\ \end{pmatrix}$$
And convert it to a triangular matrix. By Schur's Theorem (the Schur Decomposition), there is a Unitary matrix $U$ so that $U^{*}AU$ is upper triangular. Mathematica doesn't want to do it with parameters, nor does it even want to do it if I put in integers (something about infinite precision). In any case, I really need the unitary matrix, as well as the triangular matrix that results.
What I am hoping for is this: Going back the other way, I hope that for such a unitary matrix $U$, that for any (5x5) triangular matrix $T$, $U^{*}TU$ is "upside-down triangular," even if the entries end up in different spots. Proving that would be enough, even if it can be proven abstractly without finding a form for $U$.
There exists no unitary matrix $U$ such that for every triangular matrix $T$, $UTU^*$ is upside-down triangular.
To see this, it suffices to note that the identity matrix $I$ is triangular, but that $UIU^* = UU^* = I$ will fail to be upper-triangular for any unitary matrix $U$.