In order to get to the parts I am stuck at, I will add the examiners' solutions to each subquestion, which is needed to get to the subquestion that I am querying.
The following is a bizarre question from a highly ranked university quantum mechanics exam:
Consider the quantity
$$\hat{\mathcal{O}}_{mn}(t)=\int {u_m}^{*}(x,t)\hat{\mathcal{O}}\,u_n(x,t)dx$$ for some operator $\hat{\mathcal{O}}$ which has no explicit time-dependence, where $u_n(x,t)$ and $u_m(x,t)$ are eigenstates of $\hat{H}$ at time $t$.
Write this expression in terms of the eigenstates $u_m(x)$ and $u_n(x)$ at time $t=0$
In the second line I believe that there is a mistake and $\int {u_m}^{*}(x,t)\hat{\mathcal{O}}\,u_n(x,t)dx$ should be $\int {u_m}^{*}(x,0)\hat{\mathcal{O}}\,u_n(x,0)dx$ since all the time dependence is in the exponentials out front.
Could someone please confirm or deny whether this is indeed correct?
What is the time-derivative? $$\frac{d}{dt}\hat{\mathcal{O}}_{mn}(t)?$$
Consider the operator $\hat{\mathcal{O}} = [\hat{A},\hat{H}]$ and find $\hat{\mathcal{O}}_{mn}(t)$ for this case. Find a relationship between $\frac{d \hat{A}_{mn}}{dt}$ and $\hat{\mathcal{O}}_{mn}(t)$.
In the first integral there may be a typo as I think that $$\int {u_m}^{*}(x)[\hat{A},\hat{H}]u_n(x,t)dx$$ should be $$\int {u_m}^{*}(x)[\hat{A},\hat{H}]u_n(x)dx$$
I think that $$\hat{\mathcal{O}}_{mn}(0)=(E_n-E_m)\langle{\hat{A}\rangle}$$ when the examiner writes "Comparing to the above" in the last line, I presume the part that is being compared is $$\frac{d}{dt}\hat{\mathcal{O}}_{mn}(t)=-\frac{i}{\hbar}\left(E_n-E_m \right)\hat{\mathcal{O}}_{mn}(t)$$
I can't figure out how to derive the relationship $$\frac{d}{dt}\hat{A}_{mn}(t)=-\frac{i}{\hbar}[\hat{A},\hat{H}]_{mn}\tag{1}$$ as I don't even think it is correct since the question asked to find $\color{red}{\hat{\mathcal{O}}_{mn}(t)}\,\color{red}{\text{for this case.}}$
It also asked for $\color{red}{\text{a relationship between}}\,$ $\color{red}{\frac{d \hat{A}_{mn}}{dt}}$ $\color{red}{\text{and}}\,$ $\color{red}{\hat{\mathcal{O}}_{mn}(t)}$ not $[\hat{A},\hat{H}]$.
I would really like to understand how to derive relation $(1)$, so if anyone could give me any hints or advice on how to go about this it will be much appreciated.
I'm sure you are right about all the typos, but I think the answer is essentially correct
$$ \hat{\mathcal{O}}_{mn}(0)= [\hat A, \hat H]_{mn}(0)=(E_n-E_m){\hat A}_{nm} (0)$$ so $$ \hat{\mathcal{O}}_{mn}(t)= [\hat A, \hat H]_{mn}(t)= e^{-\frac i \hbar (E_n-E_m)t }(E_n-E_m) {\hat A}_{nm} (0) \\ = (E_n-E_m) {\hat A}_{nm} (t) $$ Also, assuming $ \hat A$ has no explicit time dependence, it must satisfy $$ \frac{d}{dt}\hat{A}_{mn}(t)=-\frac{i}{\hbar} (E_n-E_m) \hat A_{mn}(t) \\ = -\frac{i}{\hbar} \hat{\mathcal{O}}_{mn}(t) $$