Is this induced action transitive? Let $G$ be a group of order $n$ and let $k$ be a smallest integer such that 
we have a injection from $G$ to $S_k$. Denote $\bar G$  as a image of $G$.
Is the action of $\bar G$ on $\{1,2,...,k\}$  transitive ?
 A: Not necessarily.  For example, if $G=\mathbb{Z}/6$, then the smallest symmetric group containing a copy of $G$ is $S_5$.  But the elements of order $6$ of $S_5$ are products of two disjoint cycles, one of length $2$ and one of length $3$.  The group generated by such an element is never transitive (it admits two orbits).
A: Not necessarily, when $G=C_6$, $k=5$ with $\overline{G} = \langle (1,2)(3,4,5) \rangle$.
More generally, if $G$ and $H$ are groups with $G \le S_k$ and $H \le S_l$, then $G \times H \le S_k \times S_l$ and that will nearly always give you the smallest embedding for $G \times H$.
A: To give a smaller example, let $G = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.  Then, in order for there to be an injection $\iota: G \to S_k$, we require that $S_k$ have at least $3$ commutinge elements of order $2$.  The least such $k$ is $k=4$.  But then, we can choose $\iota$ so that $\overline{G} = \{(1), (1\;2), (3\;4), (1\;2)(3\;4)\}$.  Then, the orbit of $1$ under the action of $\overline{G}$ does not include $3$ or $4$, so $\overline{G}$ does not act transitively.

In fact, the only finite abelian groups for which $\overline{G}$ acts transitively on $\{1,2,\cdots k\}$ are $G \cong \mathbb{Z}/p^r\mathbb{Z}$ (EDIT: and, as Derek Holt points out in the comments, $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$), where $p$ is a prime.  To see why, we first observe that the minimal $k$ such that $\iota: \mathbb{Z}/p^r\mathbb{Z} \to S_k$ is injective is $k = p^r$ (since for smaller $k$, $S_k$ fails to have an element of order $k$), and that $\overline{G}$ acts on the set $\{1,2,\cdots, k\}$ by cyclically permuting its elements.  
On the other hand, if $G$ is not isomorphic to a group of the form $\mathbb{Z}/p^r\mathbb{Z}$, then we can write $G$ as
$$G \cong \prod_{i=1}^t \mathbb{Z}/p_i^{n_i}\mathbb{Z}$$
where the $p_i$ are (nont necessarily distinct) primes.  Such a $G$ can be realized as a subgroup $\overline{G}$ of $S_k$ for $k = \sum_i p_i^{n_i}$ defining $\iota$ to send the generator $g_i$ of each of the component subgroups $\mathbb{Z}/p_i^{n_i}\mathbb{Z}$ to a disjoint $p_i^{n_i}$ cycle.  In fact, this is the minimum such $k$, since in order that $\iota(g_i)\iota(g_j) = \iota(g_j)\iota(g_i)$ we require that $\iota(g_i)$ and $\iota(g_j)$ are disjoint.  But then, the orbit of any element contained in the cycle of $\iota(g_i)$ will be the elements contained in $\iota(g_i)$, so for $t > 1$ the set $\{1,2,\cdots, k\}$ has multiple orbits under the action of $\overline{G}$.
