Say we have a four-dimensional spherically symmetric $\mathfrak{su}(N)$ gauge potential in standard Schwarzschild co-ordinates which can be written
\begin{equation} \mathcal{A}=Adt+Bdr+\frac{1}{2}(C-C^\dagger)d\theta-\frac{i}{2}\left[(C+C^\dagger)\sin\theta+D\cos\theta\right]d\phi, \end{equation}
where $A$ and $B$ purely imaginary, traceless and diagonal matrices that in general depend on $t$ and $r$, $D$ is a real constant traceless and diagonal matrix, $C$ is a complex upper-triangular matrix depending on $t$ and $r$, and $C^\dagger$ is its Hermitian conjugate. (See the work of H. P. Kunzle, Comm. Math. Phys 162, 1994).
Now I am reliably informed that there exists a residual $\mathfrak{su}(N)$ Lie algebra degree of freedom, and so if $\mathfrak{g}(t,r)$ is an $\mathfrak{su}(N)$-valued diagonal matrix, the following gauge transform leaves $\mathcal{A}$ invariant but alters the matrices in the potential, as
\begin{equation} \begin{split} A&\rightarrow A+\mathfrak{g}^{-1}\dot{\mathfrak{g}},\\ B&\rightarrow B+\mathfrak{g}^{-1}\mathfrak{g}^\prime,\\ C-C^\dagger&\rightarrow\mathfrak{g}^{-1}(C-C^\dagger)\mathfrak{g},\\ C+C^\dagger&\rightarrow\mathfrak{g}^{-1}(C+C^\dagger)\mathfrak{g},\quad\quad(1) \end{split} \end{equation}
where $\dot{f}=\frac{df}{dt}$ and $f^\prime=\frac{df}{dr}$. However I'm having trouble finding with why this is the case. Here is as far as I've got:
We may perform a gauge transform on the potential $\mathcal{A}$:
\begin{equation} \mathcal{A}\rightarrow\mathfrak{g}\mathcal{A}\mathfrak{g}^{-1} \end{equation}
which given that $\mathfrak{g}$, $A$, $B$ and $D$ are diagonal, yields
\begin{equation} \mathcal{A}=Adt+Bdr+\frac{1}{2}\mathfrak{g}(C-C^\dagger)\mathfrak{g}^{-1}d\theta-\frac{i}{2}\left[\mathfrak{g}(C+C^\dagger)\mathfrak{g}^{-1}\sin\theta+D\cos\theta\right]d\phi. \end{equation}
Now applying $(1)$ gives
\begin{equation} \begin{split} \mathcal{A}=&Adt+\mathfrak{g}^{-1}\dot{\mathfrak{g}}dt+Bdr+\mathfrak{g}^{-1}\mathfrak{g}^\prime dr+\frac{1}{2}(C-C^\dagger)d\theta-\frac{i}{2}\left[(C+C^\dagger)\sin\theta+D\cos\theta\right]d\phi\\ =&Adt+\mathfrak{g}^{-1}d\mathfrak{g}+Bdr+\mathfrak{g}^{-1}d\mathfrak{g}+\frac{1}{2}(C-C^\dagger)d\theta-\frac{i}{2}\left[(C+C^\dagger)\sin\theta+D\cos\theta\right]d\phi. \end{split} \end{equation}
What am I to make of the extra term which appears to be $2\mathfrak{g}^{-1}d\mathfrak{g}$? Is it somehow zero for some reason I can't see, or have I made an error earlier on? Any help is greatly appreciated.
Haha ok I've done that thing where I figure out the answer to my own question 10 mins after asking it. May as well give the solution, it might help others.
Of course, my transform
\begin{equation} \mathcal{A}\rightarrow\mathfrak{g}\mathcal{A}\mathfrak{g}^{-1} \end{equation}
is faulty - the correct expression (letting $\mathcal{A}\equiv\mathcal{A}_\mu dx^{\mu}$) is given by
\begin{equation} \mathcal{A}_\mu\rightarrow\mathfrak{g}\mathcal{A}_\mu\mathfrak{g}^{-1}-(\partial_\mu\mathfrak{g})\mathfrak{g}^{-1}. \end{equation}
This causes the problem terms to vanish, as this instead yields (noting $\mathfrak{g}$ is diagonal)
\begin{equation} \begin{split} \mathcal{A}\rightarrow&Adt-\dot{\mathfrak{g}}\mathfrak{g}^{-1}dt+Bdr-\mathfrak{g}^\prime\mathfrak{g}^{-1}dr+\frac{1}{2}\mathfrak{g}(C-C^\dagger)\mathfrak{g}^{-1}d\theta\\&-\frac{i}{2}\left[\mathfrak{g}(C+C^\dagger)\mathfrak{g}^{-1}\sin\theta+D\cos\theta\right]d\phi\\ \rightarrow& Adt+\mathfrak{g}^{-1}\dot{\mathfrak{g}}dt-\dot{\mathfrak{g}}\mathfrak{g}^{-1}dt+Bdr+\mathfrak{g}^{-1}\mathfrak{g}^\prime dr-\mathfrak{g}^\prime\mathfrak{g}^{-1}dr+\frac{1}{2}(C-C^\dagger)d\theta\\ &-\frac{i}{2}\left[(C+C^\dagger)\sin\theta+D\cos\theta\right]d\phi\\ \rightarrow&\mathcal{A}. \end{split} \end{equation}