Question: Let $A$ be any $2\times 2$ matrix in $\mathbb{C}^2$ and let
$$X=\begin{pmatrix} \cos(\theta) & -e^{i\lambda}\sin(\theta)\\ e^{i\phi}\sin(\theta) & e^{i(\lambda+\phi)}\cos(\theta) \end{pmatrix}$$
(the most general $2\times 2$ unitary matrix). Are there angles $(\theta,\lambda,\phi)$ (where each of these angles live in the interval $[0,2\pi)$) such that
$$\text{det}([A,X])=\left|A X-XA\right|=0$$
other than the trivial case $(0,\lambda,2\pi-\lambda)$.
Previous Work: First, the trivial case is ignored because otherwise $X=I$ would always force a zero determinant. I have also proved (by exhaustion) this conjecture for the cases when two or more elements of $A$ are zero and it always works. Additionally, I have done numerical tests and can always find non-trivial angles which give a zero determinant. Of important note: it is possible that for some matrices $A$ the resulting matrix $[A,X]=AX-XA\neq \mathbf{0}$ (for any angles other than the trivial angles). However, in all these cases (at least that I have found) there are still angles which give a zero determinant corresponding to a zero eigenvalue with multiplicity one (as opposed to $\mathbf{0}$ which has a zero eigenvalue of multiplicity two).
Motivation: This is a continuation of my question here and the motivation remains the same. Basically, I have simplified that question down (and slightly generalized it) to the question presented in this post.
I feel that there should be a simple way to prove this, but for the life of me I cannot show it in generality.
Choose a nonzero eigenvector $v$ of $A$; this is guaranteed to exist. Pick also a second vector $v'$ orthogonal to $v$. We have $Av=\lambda v$ with some complex $\lambda$, by definition. Define $X$ by $Xv=z v$, $X v'=z' v'$, where $z,z'$ are complex numbers of unit modulus. This $X$ is then unitary and $[A,X]v=0$, so $[A,X]$ has zero determinant. If we assume that $ A $ is hermitian or more generally normal, we can even choose an orthonormal basis of $A$ and construct unitary $X$ such that $[A,X]=0$.