Consider two matrices $A,B \in \mathfrak{su}(N)$ which are both diagonal in the standard basis and non-zero.


If we consider the new matrix $\tilde{B} := FBF^{\dagger}$ where $F$ is the `quantum' fourier transform matrix:


$$F\_N = \frac{1}{\sqrt{N}} \begin{bmatrix}
1&1&1&1&\cdots &1 \\
1&\omega&\omega^2&\omega^3&\cdots&\omega^{N-1} \\
1&\omega^2&\omega^4&\omega^6&\cdots&\omega^{2(N-1)}\\ 1&\omega^3&\omega^6&\omega^9&\cdots&\omega^{3(N-1)}\\
\vdots&\vdots&\vdots&\vdots&&\vdots\\
1&\omega^{N-1}&\omega^{2(N-1)}&\omega^{3(N-1)}&\cdots&\omega^{(N-1)(N-1)}
\end{bmatrix}$$
(where $\omega=\exp(2i\pi/N)$, see the [wikipedia page on Quantum Fourier Transform](https://en.wikipedia.org/wiki/Quantum_Fourier_transform))


what are the possibilities for the Lie algebra generated (i.e. all linear combinations of all bracket expressions) as $\left<A, \tilde{B}\right>\_{Lie}$? I.e. what algebras can be generated this way by varying $B$ while keeping it diagonal.


Specifically, will the Lie group associated to whatever algebra is generated contain a subgroup isomorphic to any non-trivial Clifford group?



What is this Lie algebra?
I can't see your question having a completely general answer. In the 'generic' case, I think the subalgebra generated by $A$ and $\tilde{B}$ is just $\mathfrak{su}(N)$.


To see this, first of all note that $\tilde{B}$ is of the form $$\begin{bmatrix} 0 & b\_1 & b\_2 & \dots & b\_{N-1} \\ b\_{N-1} & 0 & b\_1 & \dots & b\_{N-2} \\
b\_{N-2} & b\_{N-1} & 0 & \dots & b\_{N-3} \\
\vdots & \vdots & \vdots & \dots & \vdots \\
b\_{1} & b\_{2} & b\_{3} & \dots & 0 \end{bmatrix}$$
with $b\_{N-i}=-\overline{b\_i}$ for $1\leq i\leq N-1$, and all such matrices are possible.
Let $A$ have diagonal entries $a\_1{\rm i},\ldots ,a\_N{\rm i}$ (where the $a\_i$ are real and sum to zero). In the generic case, the square differences $(a\_i-a\_j)^2$ for $i<j$ are all distinct.
Then the subspace of elements $x$ of $\mathfrak{su}(N)$ satisfying $[A,[A,x]]=-(a\_i-a\_j)^2 x$ is two-dimensional, spanned by $e\_{ij}-e\_{ji}$ and ${\rm i}(e\_{ij}+e\_{ji})$, where $e\_{ij}$ denotes the matrix with $1$ in the $(i,j)$ position, and $0$ elsewhere.


Generically, the $b\_i$ are all non-zero, and hence $\tilde{B}$ is supported on this subspace of elements satisfying $[A,[A,x]]=-(a\_i-a\_j)^2 x$, specifically the component of $x$ is $$b\_{j-i}e\_{ij}-\overline{b\_{j-i}}e\_{ji}=\beta(e\_{ij}-e\_{ji})+\gamma{\rm i}(e\_{ij}+e\_{ji})$$ where $\beta,\gamma$ are real (and at least one is non-zero). Since the $(a\_i-a\_j)^2$ are distinct, it follows that $\beta(e\_{ij}-e{ji})+\gamma{\rm i}(e\_{ij}+e\_{ji})$ belongs to $\langle A,\tilde{B}\rangle$.
Now since $$[A,\beta(e\_{ij}-e\_{ji})+\gamma{\rm i}(e\_{ij}+e\_{ji})]=(a\_i-a\_j)(-\gamma(e\_{ij}-e\_{ji})+\beta{\rm i}(e\_{ij}+e\_{ji}))$$ and since we cannot have $\gamma/\beta=-\beta/\gamma$, then $\langle A,\tilde{B}\rangle$ contains both $e\_{ij}-e\_{ji}$ and ${\rm i}(e\_{ij}+e\_{ji})$.


Finally, $[e\_{ij}-e\_{ji},{\rm i}(e\_{ij}+e\_{ji})]=2{\rm i}(e\_{ii}-e\_{jj})$ is also in $\langle A,\tilde{B}\rangle$.


