@Steve: Of course (lots of head smacking).

It would be enough to (locally) find a regular sequence consisting of invariants as then $k[X]$ would be free over the subring generated by the sequence and so would $k[X]^G$ being a direct factor of it. It seems that for a polynomial ring the coefficients of the characteristic polynomial of a general linear polynomial might do the trick.

@Veen: Notice that I said invertible scalar, the scalars referring to the base $S$. I.e., the functor associates to $S$ the isomorphism classes of rigidified line bundles on $X\times S$.

In any case, disregarding the $K$-notation, the result is true and for the same reason the others are; $\mathbb{H}P^\infty$ is the quotient of $S^\infty$ by $Sp(1)$. The action is free and $S^\infty$ is contractible so (more or less) by definition the quotient is a $BSp(1)$.

From this you also see that it is just the Borel construction (associated if you prefer to geometric realisations).

If you change the rigidification by an invertible scalar you get an isomorphic rigidification by letting the scalar act on $L$.

Note that in the genus $1$ case you do not recover the elliptic curve from its Tate modules. If you twist a (CM-)curve by a rank $1$ projective module over its endomorphism ring then the result will have the same Tate modules but only be isomorphic to the curve if the module is free. The same in some sense holds true in higher genus only that the twists may not be Jacobians. Hence it is crucial to look at the fundamental group and not its abelianisation.

@Theo: You are absolutely right, I am glad for the qualifications I put in... I leave my comment in case someone else makes the same mistake or more precisely "enom till straff, androm till varnagel" ("punishment for one, warning to others" a phrase used in old Swedish laws).

To me it seems that the $G$-representations in $\mathrm{SVect}$ consist of a graded supervector space together with a differential of superdegree $1$ and internal degree $1$. Such a graded vector spaces can be thought of as a graded vector space where the even vectors of degree $n$ correspond to new degree $2n$ and the odd vectors of degree correspond to new degree $2n+1$. In the new degree the differential is of degree $1$. This looks like the category of complexes (with the Koszul rule built into the monoidal structure as it should).

@Veen: $\mathrm{Pic}^\tau$ is defined by the condition that the line bundle lie in the $\mathrm{Pic}^\tau$ of each fibre and the same i strue for $\mathrm{Pic}^0$ (though not by definition).

It seems to me that the same argument as you gives that all automorphisms for finite extensions of $\mathbb Q_p$ are continuous in the $p$-adic topology (and hence are the identity on $\mathbb Q_p$.