Graph automorphism group Let $A_w$ denote such set of positive integer $n$ that: for any two permutations $\pi_0,\pi_1\in S_n$, if $\pi_1$ is not a power of $\pi_0$, then there exists a (labeled non oriented) graph $G$ of size $n$ such that $\pi_0\in Aut(G)$ and $\pi_1\notin Aut(G)$.
Let $A_s$ denote such set of positive integer $n$ that: for any permutation $\pi \in S_n$, there exists a (labeled non oriented) graph $G$ of size $n$ with $Aut(G) = \langle\pi\rangle$. (Pointed by @Ycor.)
Q1: Is $A_w$ or even $A_s$ infinite?
Q2: Does $A_w$ or even $A_s$ contain almost all positive integers?
Q3: What are the smallest elements in these two sets?
Is there any known results? (Papers on this subjects are welcomed.)
 A: Peter and YCor already gave a counterexample, so this answer is just some additional commentary. I'll ignore loops for simplicity. If $\varGamma$ is a permutation group on $\lbrace 1,\ldots, n \rbrace$, then the 2-closure $\varGamma_2$ of $\varGamma$ is the largest subgroup of $S_n$ such that the orbits of $\varGamma_2$ on the set of unordered pairs $\lbrace v,w\rbrace$ are the same as the orbits of $\varGamma$ on the set of unordered pairs. Clearly $\varGamma\le\varGamma_2$. (Google "2-closure" and "orbital" for tons of references and more information.)
Now, the key observation is that every automorphism group of a graph is 2-closed; i.e. it is equal to its own 2-closure. So to find counterexamples to your question just find any $\pi_0$ such that $\langle \pi_0\rangle$ is not 2-closed. YCor's example is a permutation with one cycle: for $n\ge 3$ and $\pi_0=(1\,2\,\ldots\,n)$, the 2-closure of $\langle \pi_0\rangle$ also contains an additional $n$ elements of order 2 (think of the reflections of a regular polygon about a line through the centre).
(Incidentally, many sources define "2-closed" using ordered pairs rather than unordered pairs. For undirected graphs we need unordered pairs.)
ADDED.
I'll start but not yet complete a full description of the set $F(n)$ defined by Peter.  To recap:  $F(n)$ is the set of all pairs $(\pi_0,\pi_1)$ of permutations in $S_n$ such that every $n$-vertex graph $G$ that has $\pi_0$ as an automorphism also has $\pi_1$ as an automorphism. Describing $F(n)$ is equivalent to describing the groups $\Gamma_{\pi_0}=\lbrace \pi_1 \mid (\pi_0,\pi_1)\in F(n)\rbrace$.  ($\Gamma_{\pi_0}$ is a group because the intersection of groups is a group.)
The cases $n=1$ and $n=2$ are trivial and so is the case $n\ge 3, \pi_0=(1)$; I'll leave them for the reader. Assume that $n\ge 3 $ and that $\pi_0$ is not the identity.
Two obvious facts, the first by definition the second by symmetry: (1) $\langle \pi_0\rangle\le \Gamma_{\pi_0}$. (2) For any $\gamma\in S_n$, $\Gamma_{\pi_0^\gamma}=\Gamma_{\pi_0}^\gamma$, where exponentiation denotes conjugacy.Fact (2) says that we only need be concerned with the cycle structure of $\pi_0$.
So take $\pi_0\in S_n$ and let the cycles $C_1,C_2,\ldots,C_k$ of $\pi_0$ have lengths $\ell_1\ge \ell_2\ge\cdots\ge\ell_k$.
For $1\le j\le k$, let $D_j$ be the unique dihedral group of degree $\ell_j$ that contains $C_j$. (Think of $C_j$ as the rotational symmetries of a regular $\ell_j$-gon and $D_j$ as the full symmetry group including reflections.) For small orbits with $\ell_j\le 2$, take $D_j=C_j$. Now define $\Gamma^*_{\pi_0}= D_1\oplus \cdots \oplus D_k$. By a $t$-gon we mean a single vertex if $t=1$, a single edge if $t=2$ and a (graph) cycle of length $t$ if $t\ge 3$. If we insert a $t$-gon into a $t$-cycle of $\pi_0$, we'll assume that the $t$-cycle is an automorphism of the $t$-gon.
Lemma 1. $~\Gamma_{\pi_0}\le \Gamma^*_{\pi_0}$.
Proof. For any $\ell_j\ge 2$, take a graph with an $\ell_j$-gon in $C_j$ and isolated vertices elsewhere. The automorphism group fixes $C_j$ setwise, so $\Gamma_{\pi_0}$ does too. If there are any 1-cycles, connect them into a path and connect one end of the path to some $t$-cycle ($t\ge 2$) into which insert a $t$-gon. No more edges. The automorphism group fixes each of the 1-cycles, so $\Gamma_{\pi_0}$ does too. Thus we have proved that $\Gamma_{\pi_0}$ fixes the cycle partition of $\pi_0$ cell-wise. Now take a graph in which $C_j$ contains an $\ell_j$-gon for every $j$ and no extra edges. The subgroup of its automorphism group that fixes the cycle partition of $\pi_0$ cell-wise is $\Gamma^*_{\pi_0}$. This completes the proof.
$\square$
To complete the determination of $\Gamma_{\pi_0}$, we have to determine which elements of $\Gamma^*_{\pi_0}$ preserve all possible joins between different cycles. This is slightly complicated because it depends on arithmetic properties: the number of edges between $C_i$ and $C_j$ is divisible by the least common multiple of $\ell_i$ and $\ell_j$. On the bright side, the solution for two cycles should imply the solution for any number of cycles. I need to think of how to do it smoothly.
