It seems that one can reduce the problem to the analogous one for *rooted* trees $(T,o)$, which is more tractable, as rooted trees are more rigid.  Here, the "root" $o$ is a choosen extremal vertex of the tree $T$, and all extremal vertices of $T$ different form $o$ are "leaves". An automorphism of $(T,o)$ is an automorphism of $T$ that fixes $o$. 

**1.** let's consider the analogous poset $RATF$ of all finite, automorphism-free rooted trees  (including the trivial one-vertex tree), with the analogously defined order. I claim that the only minimal element is the trivial rooted tree. Otherwise,  by contradiction, let $(T,o)$ be a minimal rooted tree, ad assume it has a minimal number of vertices among all non-trivial minimal rooted trees. Let $x\in T$ be a leaf of its, and let $y$ be its father. There exists   a non-trivial automorphism $\phi$ of $(T\setminus\{x\} ,o)$ that does not fix $y$ --otherwise, it could be extended to a nontrivial automorphism of $(T,o)$. In particular, $y\neq o$.

Let  $z$ the wedge of $x$ and $\phi(y)$ (their younger common progenitor), and let $C$ be  the connected component  of $x$ in $T\setminus\{z\}$. Clearly, $o\notin C$. Then, $(T\setminus C, o)$ is a non-trivial tree ($\phi(y)$ is a leaf of it). It has no non trivial automorphisms, otherwise one of them could be extended to $(T,o)$. Moreover, for any $w\in T\setminus C$ there exists a  nontrivial automorphisms $\psi$ of $(T\setminus\{w\},o)$ by the assumed minimality of $(T,o)$: the key point is that $C$ is invariant for $\psi$, so that by restriction $\psi$ gives a nontrivial automorphism of  $\big((T\setminus C\\big )\setminus \{w\}, o)$, showing that $(T\setminus C, o)$ is still a minimal non-trivial rooted tree, a contradiction.

**2.** Let's $T$ be a minimal element of $\mathcal {AFT}$: let's show it is $E_7$. For any leaf $x$ of $T$, let's denote $d_T(x) > 0$ the distance from the closest bifurcation (so e.g. in $E_7$ these distances are $1,2,3$ for the three leaves). By looking at the minimum of $d$ among all leaves, we see that there is a leaf with $d_T(x)=1$, so that it's father, say $y$, is not a leaf in $T\setminus\{x\}$. Indeed, otherwise we may consider  an automorphism $\phi$ of $T\setminus\{x\}$ that does not fix $y$: then $ \phi(y) $ is a leaf with $ d_T(\phi(y)) =d_T(\phi(y)) < d_T(x)$.   

Finally, consider $ T\setminus\{x\}$. It has a non trivial automorphism which is an involution, because otherwise one could make a non trivial automorphism of $T$, freezing part of it. So  $ T\setminus\{x\}$ is made by two isomorphic copies of a rooted tree $(S,o)$ and $(S',o')$ either identifying the roots or joining them by a further edge (according with the parity of the cardinality of $T$); moreover, $T$ is obtained planting  $x$ somewhere to break the symmetry. Now the conclusion is erached this way: if $S$ is the two edges tree, then $T$ is $E_7$; otherwise, by similar arguments as before, $(S,o)$ is also a minimal element as a rooted tree, leading to a contradiction.