$N$ is prime and $E$ is of course an elliptic curve defined over $\mathbb {Q}$.
My attempt so far:
Consider the Galois action of $G = Gal(\mathbb{Q}(\zeta_N)/\mathbb{Q})\cong (\mathbb{Z}/N\mathbb{Z})^\times$ on $E[N]$ and the Weil pairing $e_N$. Let $\sigma$ be a generator of the Galois group. Let $Q$ be a point of $E[N]$ such that $e_N(P,Q) = \zeta_N$.
$\zeta_N^\sigma = e_N(P,Q)^\sigma = e_N(P,Q^\sigma)$
Applying $N-1$ times the automorhism $\sigma$ one obtains $N-1$ distinct values for $\zeta_N^\sigma$, that is, all the powers of $\zeta_N$ except $1$. Therefore, all of the points $Q^\tau$ are all different for all $\tau\in G$.
I would like to show that $Q^\sigma \in \langle Q \rangle$.
Suppose $\zeta_N^\sigma = \zeta_N^a$ for some integer $a$. Then from the linearity of the Weil pairing one has
$$ \zeta_N^a = e_N(P,Q)^a = e_N(P,[a]Q) $$
$$ e_N(P,Q^\sigma - [a]Q) = 1 $$
therefore
$$ Q^\sigma - [a]Q \in \langle P \rangle $$
We could use the argument for deriving the expression above $N-1$ times, and since $\sigma^{N-1} = \sigma$:
$$ Q^\sigma - [a(N-1)]Q = Q^\sigma - [a][-1]Q = Q^\sigma + [a]Q \in \langle P \rangle $$
$$\implies [2]Q^\sigma \in \langle P \rangle $$
which I guess should imply $Q^\sigma \in \langle P \rangle$ (notice $[2]Q^\sigma$ and $Q^\sigma$ generate the same cyclic group).
We may assume $N > 2$, since the Galois action is trivial on both $E[2]$ and $\mu_2 =\{\pm 1\}$ and the statement is clear.
The point $Q$ might not quite be the generator you want. Instead, note that if $e_N(P, Q) = \zeta_N$ then you also have $e_N(P, mP + Q) =\zeta_N$ for any integer $m$, since the Weil pairing is alternating. The points $P$ and $Q$ generate $E[p]$ (abstractly as a group), so letting $\sigma$ generate $\text{Gal}(\Bbb{Q}(\zeta_N)/\Bbb{Q})$, you can always write $Q^\sigma = rP + sQ$ for some $s, r \in \Bbb{Z}/N\Bbb{Z}$. Moreover, $s \neq 1$, since we need to have $$e_N(P^\sigma, Q^\sigma) = \zeta_N^{\sigma} \neq \zeta_N = e_N(P, Q) = e_N(P,rP + Q).$$ (Here we use $N \neq 2$.) Therefore we may replace $Q$ with $Q' = Q + \frac{r}{s - 1}P$ (division modulo $N$) so that $$(Q')^\sigma = Q^{\sigma} + \frac{r}{s - 1}P^{\sigma} = rP + sQ + \frac{r}{s - 1}P = sQ'.$$ So this choice of $Q'$ does what you want while still satisfying $e_N(P, Q') = \zeta_N$.