Bounded operators on the Stinespring representation space Let $A$ be a $C^*$-algebra and let $\phi:A\to B(H)$ be a completely positive map. The Stinespring representation theorem constructs a representation of $A$ on a Hilbert space $K$, which is constructed as follows. Define $K_0=A\otimes H$ with the inner product
$$\langle a\otimes v, b\otimes w\rangle_{\phi}= \langle \phi(b^*a)v,w\rangle_H,$$
extended linearly. Then $K$ is the closure of $K_0/\{z:\langle z,z\rangle_\phi=0 \}$ in the norm induced by the inner product $\langle \cdot,\cdot\rangle_\phi$.
Given a unitary linear operator $T$ on $H$ we can consider an operator $S$ on $K$ induced by 
$$S(a\otimes v)=a\otimes (Tv).$$
My question is under what conditions on $T$ is $S$ well-defined and bounded? 
In general, I don't want to assume that $T$ commutes with the range of $\phi$.
 A: Rather than use the proof of Stinespring, I prefer to work with the uniqueness part of the result.  Given $\phi$ we can find a Hilbert space $K$, a linear map $V:H\rightarrow K$ and $\pi:A\rightarrow B(K)$ a $*$-representation with $\phi(a) = V^*\pi(a)V$.  Under the assumption that $\{ \pi(a)V\xi : a\in A, \xi\in H \}$ is linearly dense in $K$, the triple $(K,\pi,V)$ is unique up to unitary equivalence.
Now, a very simple example is when $A=\mathbb C$.  Let $\phi(1) = R^*R$ for some $R\in B(H)$.  Then we find that $K$ is the range space of $R$, and $V$ is the corestriction of $R$ to a map $H\rightarrow K$.  $\pi$ is of course the trivial rep of $\mathbb C$ on $K$.
So in this setting you are asking, given $T\in B(H)$, does there exist $S_0$ on $K$ with $S_0 R = R T$.  We can extend $S_0$ to $H$, so this is equivalent to asking:

Given $R,T\in B(H)$ when is there $S\in B(H)$ with $SR = RT$?

I don't see how to "solve this", unless you have some sort of "commutes" condition.  (By "solve this" I mean give some sort of non-trivial characterisation.)
Given that the easiest case seems intractable, I'm not sure the general case will be possible, unless you have further conditions.
