What you need is an example of a subgroup $G\subset A_n$ such that the normalizer of $G$ in $S_n$ is contained in $A_n$. If every automorphism of $G$ is inner, then it will be enough if the centralizer of $G$ in $S_n$ is contained in $A_n$. How about $n=8$ and $G$ the diagonal copy of $S_4$ in $S_4\times S_4\subset S_8$, if you know what I mean?
EDIT: If you replace this $G$ by its normal subgroup of order $4$, you get another example. This one is minimal with regard to the order of $G$ (though not with regard to $n$, as Derek's Holt's examples show).
I have come across a fairly recent result which pertains to this question. It is in this paper:
Guralnick, Robert M.; Maróti, Attila; Pyber, László, Normalizers of primitive permutation groups, Adv. Math. 310, 1017-1063 (2017). ZBL1414.20002.
An arXiv version is here. The paper examines the situation when $U$ is primitive. They show that, in all but a finite number of situations, $|N(U)/U|<n$. Indeed, they strengthen this bound if you add in a particular infinite family. The main result is this one:
Theorem: Let $U$ be a primitive subgroup of $S_n$, and let $N=N_{S_n}(U)$. Then $|N/U|< n$ unless $U$ is an affine primitive permutation group and the pair $(n, N/U)$ is one of:
$$(3^4,O^−_4(2),(5^4,Sp_4(2)),(3^8,O^−_6(2)),(3^8,SO^−_6(2)),(3^8,O^+_6(2)),(3^8,SO^+_6(2)),(5^8,Sp_6(2)),(3^{16},O^−_8(2)),(3^{16},SO^−_8(2)),(3^{16},O^+_8(2)), \textrm{ or }(3^{16},SO^+_8(2)).$$ Moreover if $N/U$ is not a section of $\Gamma L_1(q)$ when $n=q$ is a prime power, then $|N/U|< n^{1/2}\log n$ for $n≥2^{14000}$.
@YCor's comment on the original question suggests that the primitivity assumption is not too onerous. It would be interesting (but probably very hard) to try and understand what might happen when $U$ is transitive and imprimitive.
Best Answer
Yes. One way to see this is to use the fact that the smallest degree faithful permutation representations of $S_n$ for $n \ge 6$ have degrees $n$ (natural action), $2n$ (imprimitive action on cosets of $A_{n-1}$), and $n(n-1)/2$ (action on unordered pairs). I think this was probably proved before the classification of finite simple groups, but I am not sure. A general reference for maximal subgroups of $A_n$ and $S_n$ is
Liebeck, Martin W.; Praeger, Cheryl E.; Saxl, Jan. A classification of the maximal subgroups of the finite alternating and symmetric groups. J. Algebra 111 (1987), no. 2, 365–383.
Let's assume $n \ge 11$ (although I think your result is true for $n \ge 7$) and $k \ge n-k$, so $k \ge 6$. So the only faithful permutation representations of $S_k$ of degree at most $n$ have degree $k$, or $2k$ when $n=2k$.
The imprimitive transitive representation of $S_k$ of degree $2k$ has centralizer of order 2 in $S_{2k}$ (the order of the centralizer in the symmetric group of a transitive permutation group is equal to the number of fixed points of its point-stabilizer), so that cannot arise as your subgroup $H \cong S_k$.
Hence $H$ must have an orbit of length $k$ and act naturally on that orbit. If it had two orbits of length $k$ (with $n=2k$), then its centralizer in $S_n$ would have order 2, so it has a unique such orbit. So your subgroup $D$ isomorphic to $S_{n-k}$, which centralizes $H$, must fix that orbit, and hence fix it pointwise, and so $D$ must act naturally on the remaining points. Hence $H$ fixes all remaining points, and the result follows.