Yes, I believe so.

Sq^2: K(Z/2,1) -> K(Z/2,3) is nullhomotopic. So your argument actually shows that the first obstruction vanishes. The next obstruction lives in H^5(BZ/2, Z) and therefore automatically vanishes. Higher obstructions vanish because the map from Pic(KU) to its 3-truncation splits (the splitting coming from the map from the Clifford category into Pic(KU) ).

I believe that this essentially never exists as an infinite loop map: for any commutative ring spectrum R, multiplication by η carries the class of R[1] in pi_0 of Pic(R) to -1 in (pi_0 R)^* = pi_1( Pic(R) ), so R[1] cannot be obtained from an infinite loop map even from Z into Pic(R) unless R has characteristic 2.

The isomorphism you describe is not an isomorphism of symmetric monoidal functors. To see this, let $k^G$ denote the regular representation of $G$, regarded as commutative algebra via pointwise multiplication. If F: Rep(G) -> Vect(k) is any symmetric monoidal functor, then $F(k^G)$ will inherit the structure of a commutative algebra. In the above example, $F(k^G)$ is the space of $G$-invariants in $k'^{G}$, where $G$ is acting both by permuting the factors and by Galois symmetries.Evaluation at the identity element of $G$ determines an isomorphism of this algebra with $k'$,which is not $k^G$.

In this context, a $G$-torsor means a $k$-scheme with an action of $G$, which is ($G$-equivariantly) isomorphic to $G$ over some extension field of $k$. So if $k'$ is a Galois extension of k with Galois group $G$, then the spectrum of $k'$ is a $G$-torsor (since $k' \otimes_k k' \simeq k^G$ by virtue of the Galois assumption).