= Solution
The two oscillator families commute with each other. Their <canonical commutation relations> give
$$
[Q,b_{\mathbf p}]=+b_{\mathbf p},\qquad
[Q,c_{\mathbf p}^\dagger]=+c_{\mathbf p}^\dagger,
\qquad
[Q,b_{\mathbf p}^\dagger]=-b_{\mathbf p}^\dagger,\qquad
[Q,c_{\mathbf p}]=-c_{\mathbf p}.
$$
Both terms in the <mode expansion of a free field> therefore acquire the same factor, giving
$$
\boxed{[Q,\phi(x)]=\phi(x),\qquad [Q,\phi^\dagger(x)]=-\phi^\dagger(x).}
$$
The <Noether charge> is the generator of the internal <symmetry>. In the active-operator convention $U(\alpha)=e^{i\alpha Q}$, repeated <commutators> give
$$
\boxed{U(\alpha)\phi(x)U(\alpha)^{-1}=e^{i\alpha}\phi(x).}
$$
The sign agrees with part b. A negative sign for $[Q,\phi]$ would correspond to the opposite current convention, not to the current used here. Subtracting a multiple of the identity from $Q$ to fix the <vacuum> <Noether charge> has no effect on this generator relation.
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