**Problem**


Given a C\*-algebra $\mathcal{A}$.


Consider dynamics $\tau:\mathbb{R}\to\mathrm{Aut}(\mathcal{A})$ and $\tau':\mathbb{R}\to\mathrm{Aut}(\mathcal{A})$.  

*(More precisely, strongly continuous one-parameter groups.)*


Denote their derivations by $\delta:\mathcal{D}\to\mathcal{A}$ and $\delta':\mathcal{D}'\to\mathcal{A}$.


Then one has:
$$\delta=\delta'\implies\tau=\tau'$$
*(Here, equality is meant in terms of operators resp. maps.)*


How do I check this?


For dynamics over Hilbert spaces I would proceed by:
$$i\frac{\mathrm{d}}{\mathrm{d}t}\|\varphi(t)\|^2=\langle H\varphi(t),\varphi(t)\rangle-\langle\varphi(t),H\varphi(t)\rangle=0$$
But for the C\*-algebra case this path is not directly available.


**Disclaimer**


I hope to get a hint from here.  

*(I haven't got any respond yet from stack exchange.)*



C*-Algebras: Dynamics vs. Derivations
As was discussed in the comments, it suffices to see that $\delta$ determines $\tau$ uniquely on $\mathrm{dom}(\delta)$ which is dense in $\mathcal{A}$. Suppose that $x\_0 \in \mathrm{dom}(\delta)$. Check that $t \mapsto \tau^t(x\_0)$ is a solution to the initial value problem
\begin{align\*}
\frac{d}{dt} x(t) = \delta( x(t)) && x(0) = x\_0.
\end{align\*}
If $x$ is any solution then
$$\frac{d}{dt} \left( \tau^{-t}(x(t)) \right) = -\tau^{-t}\delta(x(t)) + \tau^{-t}\left(\delta(x(t))\right) = 0$$
and it follows that $\tau^{-t}(x(t)) \equiv x\_0$ so that $x(t) = \tau^t (x\_0)$ is the unique solution.


Basically this all works for Banach space flows too. See the answers to my [own question](https://mathoverflow.net/questions/156149/does-the-generator-of-a-1-parameter-group-of-banach-space-isometries-know-which) here.


