Let $E/K$ be an elliptic curve with a twist $E'/K$. Let $f: E_{\overline K} \to E'_{\overline K}$ be an isomorphism. Let $m(\sigma) = f^{-1}\circ f^\sigma$ be the 1-cocycle corresponding to this twist.
I would like to calculate the relation between the Galois representations on the Tate modules $T_\ell(E)$ and $T_\ell(E')$.
Let $A(\sigma)$ and $A'(\sigma)$ be the representations on the Tate modules. Also denote by $F: T_\ell(E) \to T_\ell(E')$ the map induced on the Tate modules by $f$. Similarly, let $M(\sigma)$ be the map induced by $m(\sigma)$.
From $f^\sigma(\sigma(P)) = \sigma(f(P))$, we see that $M(\sigma)(\sigma(P)) = f^{-1}(\sigma(f(P)))$. On the Tate modules, this becomes: $$F^{-1}A'(\sigma)F = M(\sigma)A(\sigma).$$
Is this correct? I am troubled by the case where $Aut(E_{\overline K})$ is not defined over $K$. In this case, I do not see why $\sigma \to M(\sigma)A(\sigma)$ should be a group homomorphism like $\sigma \to F^{-1}A'(\sigma)F$ is.
What I wrote is correct. The point is that $M(\tau\sigma) = M(\sigma)A(\tau)M(\sigma)A^{-1}(\tau)$. Using this, both sides are indeed group representations.
This is because $m(\sigma)^\tau = \tau(m(\sigma))\tau^{-1}$ and $m(\tau\sigma) = m(\tau)m^\tau(\sigma) = m(\tau)\tau(m(\sigma))\tau^{-1}$.
Note that $\sigma \to M(\sigma)$ is a 1-cocycle in $H^1(G_K, Aut(T_\ell(E)))$ where $Aut(T_\ell(E))$ is a $G_K$ representation through the adjoint representation. Therefore, we can think of $m \to M$ as the map $H^1(G_K,Aut(E)) \to H^1(G_K, Aut(T_\ell(E)))$ induced from $Aut(E) \to Aut(T_\ell(E))$.