Yes. Your notation $n$ is used for two different meanings, so let me denote by $m$ the number of summands. A homomorphism $G\to S_n$ correspond to an action of $G$ on a set with $n$ elements, which can be split into orbits $A_1,...,A_\ell$. Each orbit can can be provided with a structure of a group with a surjective homomophism $G\to A_i$. Then, the condition is precisely that the induced map $G\to \oplus_i A_i$ is injective. Note the total number of cyclic summands in the right hand side is at least $n$. If $A_i$ is not cyclic, using $ab\ge a+b$ for $a,b>1$, we see that it only decreases the sum of sizes if we split $A_i$ into two parts corresponding to a decomposition $A_i = B\oplus C$. So we may assume that all the $A_i$-s are cyclic of order a prime power. By comparing each primary component at a time, we may assume that all the $p_i$-s are a single prime $p$. The result now follows from the following standard lemma
Lemma: Let $\phi: \oplus_i \mathbb{Z}/p^{a_i} \to \oplus_j \mathbb{Z}/p^{b_j}$ be injective, with $a_1 \ge a_2 \ge ...\ge a_k$ and $b_1 \ge ...\ge b_\ell$. Then $a_i \le b_i$.
edit: As mensioned correctly by @Richard Lyons, the proof I gave for it was totally incorrect so I deleted it. Here's an hopefully less wrong proof, im really sorry for that.
Let $A\subseteq B$ be abelian finite $p$-groups. Write $B=\mathbb{Z}/p^k \oplus B'$ for $B'$ of exponent smaller than or equal $k$. Let $B\to \mathbb{Z}/p^k$ be the projection. If the composition
$A\to B\to \mathbb{Z}/p^k$ is not surjective, then $A$ is contained in $p\mathbb{Z}/p^k \oplus B$ and we can proceed by induction onthe total number of elements of $B$. Otherwise $A$ contains an element of the form $(1,b)$ for $b\in B'$. Since the exponent of $B'$ is no more than $k$, there is a map $\mathbb{Z}/p^k\to B'$ mapping $1$ to $b$, and so by applying the automorphism $(x,c)\mapsto (x,c-xb)$ we may assume that $(1,0)$ is in $A$. It follows then that $A=\mathbb{Z}/p^k \oplus A\cap B'$ and we proceed by induction on the number of summands.