Is there $t\in\operatorname{Gal}(\overline{K}/K)$ s.t. $\operatorname{rank}_{\mathbf{Z}_p}((t-1)E_{p^\infty}(\overline{K}))=1$? Let $E$ be an elliptic curve defined over $\mathbb{Q}$, let $$K:=\varinjlim_{k\in\mathbb{Q}[\mu_{p^\infty}]} \mathbb{Q}\left[\mu_{p^\infty},k^{1/p^\infty}\right]$$
and $G:=\operatorname{Gal}(\overline{K}/K)$. Suppose that $E_p(K)=0$.
Question: Is there always a $\tau\in G$ so that
$$\operatorname{rank}_{\mathbf{Z}_p}((t-1)E_{p^\infty}(\overline{K}))=1$$
 A: I think the answer is "yes" in the case that $E$ doesn't have complex multiplication.  In that case, Serre's Open Image Theorem says that the image of Galois in open in $\mathrm{GL}(T_{p}(E))$, and therefore, the image of $\mathrm{Gal}(\bar{\mathbb{Q}} / \mathbb{Q}(\mu_{p^\infty}))$ is open in $\mathrm{SL}(T_{p}(E))$ -- call it $H$.  Now $K$ is abelian over $\mathbb{Q}(\mu_{p^\infty})$, so the image of $\mathrm{Gal}(\bar{K} / K)$ is in turn a (normal) subgroup of $H$ with abelian quotient.  But since the Lie algebra of $H$ is all of $\mathfrak{sl}_{2}(\mathbb{Q}_{p})$, which is semisimple, any Lie subgroup of $H$ with abelian quotient must have the same Lie algebra.  So the image of $\mathrm{Gal}(\bar{K} / K)$ is open in $H$ and thus open (and of finite index) in $\mathrm{SL}(T_{p}(E))$. Thus, the image of $\mathrm{Gal}(\bar{K} / K)$ contains a (nonzero) power of a transvection, i.e. a matrix of the form $\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}$ for some $a \neq 0$ with respect to a suitable basis.
In the case that $E$ has complex multiplication, I suspect I have a counterexample.  Take $E: y^{2} = x^{3} - 2$ and $p = 2$.  This has complex multiplication by $\mathbb{Q}(\zeta_{3})$, where $\zeta_{3}$ a primitive cube root of unity, and its order-$2$ points are the points $(\zeta_{3}^{i}2^{1/3}, 0)$ for $i = 1, 2, 3$.  Clearly the image of $\mathrm{Gal}(\bar{K} / K)$ is isomorphic to $\mathbb{Z} / 3\mathbb{Z}$ (since $\zeta_{3} \in K$ but $2^{1/3} \notin K$), so it doesn't fix any of the points in $E[2]$.  However, I think it shouldn't be too hard to show that the image of Galois can't contain a power of a transvection in this case.  (In fact, it should be contained in the image of the endomorphism ring, which is an order in $\mathbb{Z}[\zeta_{3}]$, and note that the endomorphism corresponding to $\zeta_{3}$ must act like $\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}$ or $\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ modulo $2$.)
