(Sorry for my bad English.)
Let $k$ be a field, and $\ell$ be a prime not equal to characteristic of $k$.
Moreover $E$ be an elliptic curve over $k$, $\alpha:E\to E$ be an automorphism, and $G\subset E(\overline{k})$ be subgroup of order $\ell$. Then $\alpha$ induces automorphism $E/G\cong E/\alpha(G)$.
I want to know if this inverse is true. That is, if $G_1, G_2\subset E(\overline{k})$ be subgroups of order $\ell$ s.t. $E/G_1\cong E/G_2$ , then is there automorphism $\alpha:E\to E$ s.t. $\alpha(G_1)=G_2$?
I don't think this is true.
Let $k = \mathbb{C}$ and let $\ell$ be a prime that is $1$ mod $4$, which means $\ell = \mathfrak{l}\overline{\mathfrak{l}}$ splits in $\mathbb{Q}(i)$.
The elliptic curve $\mathbb{C}/\mathbb{Z}[i]$ has the canonical maps to $\mathbb{C}/\mathfrak{l}^{-1}$ and $\mathbb{C}/\overline{\mathfrak{l}}^{-1}$ with kernels $\mathfrak{l}^{-1}/\mathbb{Z}[i]$, respectively $\overline{\mathfrak{l}}^{-1}/\mathbb{Z}[i]$.
These quotients are isomorphic to one another, with an isomorphism given by multiplication by $\lambda/\overline{\lambda}$ where $\lambda$ is a generator of $\mathfrak{l}$. However, that isomorphism is not effected by an automorphism of $\mathbb{C}/\mathbb{Z}[i]$ moving $\mathfrak{l}^{-1}/\mathbb{Z}[i]$ to $\overline{\mathfrak{l}}^{-1}/\mathbb{Z}[i]$.
Indeed, the only automorphisms of $\mathbb{C}/\mathbb{Z}[i]$ are $\{ \pm 1, \pm i \}$, which fix $\mathfrak{l}^{-1}/\mathbb{Z}[i]$.