Poincaré group on quantum Klein-Gordon field (C*-algebraic scenario)

on the same topic as this question, I have been trying to fool around with the free real K-G field in flat spacetime on the C*-algebraic scenario (Haag-Kastler axioms, Weyl quantization, etc).

Since I'm talking about the free (linear) Klein-Gordon field, the C*-algebra is taken as the CCR algebra generated by the unitary Weyl operators $W(f)$ (with $f$ a test function) which can be looked as the exponential of the field operators, $\exp\big( i\Phi(f)\big)$.

The action of the Poincare group usually is given for the field operators (Borchers algebra), with something as $\alpha_{(\Lambda,a)}\Phi(x) = \Phi(\Lambda x + a)$, as operator-valued distributions. Now I imagine that you can transport that to Weyl operators, something as $\alpha_{(\Lambda,a)} W(f) = \exp\big( i [\alpha_{(\Lambda,a)}\Phi](f)\big)$.

My questions are

1. Is the expression for $W(f)$ correct?
2. Does the action on the Weyl unitaries extend to a nice action on the CCR-algebra? By *-automorphisms? Is it inner or outer or what?
3. Where can I read about it? I could use a "for dummies" reference...

[EDIT: fixed the notation, as sugestion of user1504]

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0) It's weird to denote the action by $Ad$; this is usually reserved for adjoint actions. I'm going to use $\rho$.
1) Your expression is correct. Note that $(\rho\Phi)(f)$ is defined to be $\Phi(\rho f)$. In the end, we're just translating and transforming the test functions.
2) It should. I'm not 100% sure. It really ought to be an inner automorphism, since one can construct generators for the Poincare algebra from the field operators. (See Peskin & Schroder Chap 2, the discussion of Noether's theorem.) But there might be annoying technicalities, stemming from your decision to use the Weyl operators, instead of the raw $\Phi(f)$ observables.
Dear user1504, thank you for your answer. I completely agree on $Ad$, I will edit the question to change that. Let me just note that your (1) answer is probably taking the "dual" action on the test functions ($(\Lambda,a)f(x) = f(\Lambda^{-1}(x-a))$). About (2), well, that was my original problem, since it seems that the generator operators come from the unitary representation, not the adjoint one. Also, since the field operators are present only when you take "analytic" states and their GNS representation, I am not sure if one can use it freely on the C*-algebraic scenario. –  Yul Otani Nov 15 '12 at 17:36