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Let me start with some general notions, that I believe are known, for a tree $T=(V,E)$ with finite, nonempty vertex set $V$ and edge set $E$. I will assume that $T$ is a minimal element of $\mathcal{AFT}$ only in the end.

2. $\mathrm{diam}(T_x)=\mathrm{diam}(T)-1$. This means that every maximal length path of $T$ must end at $x$. In this case, I think there is a further alternative (it seemed to me not hard to prove, at least assuming $T\in \mathcal{AFT}$):

either (2a) $T$ possesses a nontrivial automorphism (which is not the case if $T\in \mathcal{AFT}$), or (2b) no vertex of the minimal path $(x^0,\dots, x^{n})$ connecting $x$ to $C(T)$ is a branching point; moreover, if $T$ is assumed to be a minimal element of $\mathcal{AFT}$, $\mathrm{diam}(T)$ is an odd number $2n+1$.

Now, assume $T$ is a minimal element in $\mathcal{AFT}$, so that we should be in case 2b. Since $(x^0,\dots, x^{n-1})$ is part of a maximum length path in $T_x$, and $ \mathrm{diam}(T_x)=2n $ is even, the center of $T_x$ is a single vertex, namely the other endpoint $y^0$ of $C(T):=\{x^0,y^0\}$. If $f_x$ denote the unique nontrivial automorphism of $T_x$, we know that $f_x(y_0)=y_0$ (it's the center of $T_x$), while $y:=f_x(x^{n-1})\neq x^{n-1} $ (otherwise $f_x$ would extend to $T$). Therefore, $ (y^0, f_x(x^0),f_x(x^1),\dots,f_x(x^{n-1}))$ is the $n$-edges path connecting $y$ to $C(T)$, and since the $x^i$ for $i\ge0$ are not branching points, this path has no branching points too, with the possible exception of $y^0$. Actually, $y^0$ must be a branching point, otherwise the path $\xi:=(x^n,x^{n-1},\dots,x^0,y^0,y^1,\dots y^n)$, which has maximal length $2n+1$ in $T$, would have no branching point at all, and therefore would be $T$ itself, what however is impossible because $T$ has no nontrivial automorphism.

Next, we may consider the automorphism $f_y$ of $T_y$. As to $C(T_y)$, it is either $\{x^0\}$ (if $\xi$ is the unique maximum length path of $T$ and $ \mathrm{diam}(T_y)=2n $) , or $C(T_y)=C(T)$, (if there are other maximum length paths in $T$ and $ \mathrm{diam}(T_y)=2n+1 $). Therefore $f_y(y_0)$ is either $x^0$, or $x^1$, or $y^0$; however, only the last case is possible, because $y_0$ is a branching points and $x^0$, or $x^1$ are not. Thus, $( f_y(y^0), f_y(y^1),\dots,f_y(y^{n-1}))$ is a path of length $n-1$ , starting from the branching point $y^0=f_y(y^0)$, without other branching points. For the same reason, $T$ must contain a family of paths emanating from $y^0$, with no branching points, of all lengths between $1$ and $n$; in particular, a leaf $z$ attached to $y^0$ (and possibly other matter). The unique involution $f$ of $T_z$ exchanges the endpoints of $C(T_z)=C(T)$ (otherwise it would be extensible to a nontrivial automorphism of $T$), therefore bijects the whole $\Gamma(x^0, T)=\Gamma(x^0, T_z)$ with $\Gamma(y^0, T_z)$. This proves that $n=2$ and $T$ is $E_7$.

[edit 01.17.2013] I filled the lacking point in the case 2, small but subtle, completing the proof, so I wrote it (even if in the meanwhile a complete proof has been posted).

Let me start with some general notions, that I believe are known, for a tree $T=(V,E)$ with finite, nonempty vertex set $V$ and edge set $E$. I will assume that $T$ is a minimal element of $\mathcal{AFT}$ only in the end.

2. $\mathrm{diam}(T_x)=\mathrm{diam}(T)-1$. This means that $x$ is an end of every maximal length path of $T$.

Now, assume $T$ is a minimal element in $\mathcal{AFT}$, so that we are in case 2. Then, $C(T)$ is an edge, i.e. $\mathrm{diam}(T)$ is an odd number $2n+1$, and no vertex of the minimal path $(x^0,\dots, x^{n})$ connecting $x$ to $C(T)$ is a branching point. Proof: consider first the case of odd diameter of $T$, where $C(T)$ is an edge. Assume by contradiction that $\Gamma(x^0, T)$ is not a single path. Then, there are in it leaves $y\neq x$. Take among them the one with minimum vector $\gamma(y,T)$ in the lexicographic order. Now, since $y\neq x$, we have $\mathrm{diam}(T_y)=\mathrm{diam}(T)$, and we can argue with $y$ like in the previous case 1. The automorphism $f_y$ of $T_y$ fixes all $x^i$ because $( f_y(x^0),\dots,f_y(x^n) )$ are an end of a maximum lenght path in $T$, so they must end at $x$, which implies $f_y(x^i)=x^i$ for $0\le i \le n$. But then, $f_y$ also fixes the path $y^i$, for the same inductive argument used in point $1$ (start with the greater index $j$ such that $x^j=y^j$ and proceed looking at the cardinality of $\Gamma(y^{j+1} , T_y)$, observing that $f_ y (y ^ {j+1} ) \neq x^{j+1} =f_y(x^{j+1} ) $ because $ y^{j+1} \neq x^{i+1}$. This is a contradiction as usual, because $f_y$ does not fix the father of $y$, as already observed. For an analog reason, the case $C(T)$ is a vertex implies that $\Gamma(x,T)$, that is the whole $T$, has no branching vertices, that is, it is a path, which however is impossible because $T$ has no nontrivial automorphism.

Conclusion of the proof: Since $(x^0,\dots, x^{n-1})$ is part of a maximum length path in $T_x$, and $ \mathrm{diam}(T_x)=2n $ is even, the center of $T_x$ is a single vertex, namely the other endpoint $y^0$ of $C(T):=\{x^0,y^0\}$. If $f_x$ denote the unique nontrivial automorphism of $T_x$, we know that $f_x(y_0)=y_0$ (it's the center of $T_x$), while $y:=f_x(x^{n-1})\neq x^{n-1} $ (otherwise $f_x$ would extend to $T$). Therefore, $ (y^0, f_x(x^0),f_x(x^1),\dots,f_x(x^{n-1}))$ is the $n$-edges path connecting $y$ to $C(T)$, and since the $x^i$ for $i\ge0$ are not branching points, this path has no branching points too, with the possible exception of $y^0$. Actually, $y^0$ must be a branching point, otherwise the path $\xi:=(x^n,x^{n-1},\dots,x^0,y^0,y^1,\dots y^n)$, which has maximal length $2n+1$ in $T$, would have no branching point at all, and therefore would be $T$ itself, what however is impossible because $T$ has no nontrivial automorphism.

Next, we may consider the automorphism $f_y$ of $T_y$. As to $C(T_y)$, it is either $\{x^0\}$ (if $\xi$ is the unique maximum length path of $T$ and $ \mathrm{diam}(T_y)=2n $) , or $C(T_y)=C(T)$, (if there are other maximum length paths in $T$ and $ \mathrm{diam}(T_y)=2n+1 $). Therefore $f_y(y_0)$ is either $x^0$, or $x^1$, or $y^0$; however, only the last case is possible, because $y_0$ is a branching points and $x^0$, or $x^1$ are not. Thus, $( f_y(y^0), f_y(y^1),\dots, f_ y(y^{n-1}))$ is a path of length $n-1$ , starting from the branching point $y^0=f_ y(y^0)$, without other branching points. For the same reason, $T$ must contain a family of paths emanating from $y^0$, with no branching points, of all lengths between $1$ and $n$; in particular, a leaf $z$ attached to $y^0$ (and possibly other matter). The unique involution $f$ of $T_z$ exchanges the endpoints of $C(T_z)=C(T)$ (otherwise it would be extensible to a nontrivial automorphism of $T$), therefore bijects the whole $\Gamma(x^0, T)=\Gamma(x^0, T_z)$ with $\Gamma(y^0, T_z)$. This proves that $n=2$ and $T$ is $E_7$.

[edit 01.15.2013]

Let me start with some general notions, that I believe are known, for a tree $T=(V,E)$ with finite, nonempty vertex set $V$ and edge set $E$. I will assume that $T$ is a minimal element of $\mathcal{AFT}$ only in the end.

2. $\mathrm{diam}(T_x)=\mathrm{diam}(T)-1$. This means that every maximal length path of $T$ must end at $x$. In this case, it turns out that there is a further alternative:

either (2a) $T$ possesses a nontrivial automorphism (which is not the case if $T\in \mathcal{AFT}$), or (2b) no vertex of the minimal path $(x^0,\dots, x^{n})$ connecting $x$ to $C(T)$ is a branching point; moreover, if $T$ is assumed to be a minimal element of $\mathcal{AFT}$, $\mathrm{diam}(T)$ is an odd number $2n+1$. This is nice and not hard to prove; I'd add the details at your request.

Now, assume $T$ is a minimal element in $\mathcal{AFT}$; so we are in case in case 2b. Since $(x^0,\dots, x^{n-1})$ is part of a maximum length path in $T_x$, and $ \mathrm{diam}(T_x)=2n $ is even, the center of $T_x$ is a single vertex, namely the other endpoint $y^0$ of $C(T):=\{x^0,y^0\}$. If $f_x$ denote the unique nontrivial automorphism of $T_x$, we know that $f_x(y_0)=y_0$ (it's the center of $T_x$), while $y:=f_x(x^{n-1})\neq x^{n-1} $ (otherwise $f_x$ would extend to $T$). Therefore, $ (y^0, f_x(x^0),f_x(x^1),\dots,f_x(x^{n-1}))$ is the $n$-edges path connecting $y$ to $C(T)$, and since the $x^i$ for $i\ge0$ are not branching points, this path has no branching points too, with the possible exception of $y^0$. Actually, $y^0$ must be a branching point, otherwise the path $\xi:=(x^n,x^{n-1},\dots,x^0,y^0,y^1,\dots y^n)$, which has maximal length $2n+1$ in $T$, would have no branching point at all, and therefore would be $T$ itself, what however is impossible because $T$ has no nontrivial automorphism.

[edit 01.15.2013] The following proof is still incomplete, but the main ideas should be useful.

Let me start with some general notions, that I believe are known, for a tree $T=(V,E)$ with finite, nonempty vertex set $V$ and edge set $E$. I will assume that $T$ is a minimal element of $\mathcal{AFT}$ only in the end.

2. $\mathrm{diam}(T_x)=\mathrm{diam}(T)-1$. This means that every maximal length path of $T$ must end at $x$. In this case, I think there is a further alternative (it seemed to me not hard to prove, at least assuming $T\in \mathcal{AFT}$):

either (2a) $T$ possesses a nontrivial automorphism (which is not the case if $T\in \mathcal{AFT}$), or (2b) no vertex of the minimal path $(x^0,\dots, x^{n})$ connecting $x$ to $C(T)$ is a branching point; moreover, if $T$ is assumed to be a minimal element of $\mathcal{AFT}$, $\mathrm{diam}(T)$ is an odd number $2n+1$.

Now, assume $T$ is a minimal element in $\mathcal{AFT}$, so that we should be in case 2b. Since $(x^0,\dots, x^{n-1})$ is part of a maximum length path in $T_x$, and $ \mathrm{diam}(T_x)=2n $ is even, the center of $T_x$ is a single vertex, namely the other endpoint $y^0$ of $C(T):=\{x^0,y^0\}$. If $f_x$ denote the unique nontrivial automorphism of $T_x$, we know that $f_x(y_0)=y_0$ (it's the center of $T_x$), while $y:=f_x(x^{n-1})\neq x^{n-1} $ (otherwise $f_x$ would extend to $T$). Therefore, $ (y^0, f_x(x^0),f_x(x^1),\dots,f_x(x^{n-1}))$ is the $n$-edges path connecting $y$ to $C(T)$, and since the $x^i$ for $i\ge0$ are not branching points, this path has no branching points too, with the possible exception of $y^0$. Actually, $y^0$ must be a branching point, otherwise the path $\xi:=(x^n,x^{n-1},\dots,x^0,y^0,y^1,\dots y^n)$, which has maximal length $2n+1$ in $T$, would have no branching point at all, and therefore would be $T$ itself, what however is impossible because $T$ has no nontrivial automorphism.

Let me start with some general notions, that I believe are classic, for a tree $T=(V,E)$ with finite, nonempty vertex set $V$ and edge set $E$. I will assume that $T$ is a minimal element of $\mathcal{AFT}$ only in the end.

For a path of length $n$ (number of edges) , $ (v_0 \dots v_n) $ in $ T $, let's define the centre of the path as the set $ \big\{v _ {\lfloor\frac{n }{2}\rfloor},v _ {\lceil \frac{n }{2}\\rceil } \big\}$, consisting of one or two vertices (thus, either the middle vertex, if $n$ is even, or the middle edge, if $n$ is odd). Given two paths, there is a third one including the centres of both, and one endpoint of each. As a consequence, all paths of maximum length in a tree share the same centre, that we can therefore refer to as "centre of the tree", $C(T):=\{v,v'\}$ (so this notation allows that $C(T)=\{v\}=\{v'\}$, a singleton, precisely whenever the diameter of $T$ is an even number, as remarked).

The centre determines a natural genealogy order in $T$; in particular, to any vertex $v$ we can associate the progeny of $v$, $\Gamma(v,T)$, the set of all vertices $x$ such that the minimal path from $x$ to the centre passes by $v$. Thus, this reduces to $\{v\}$ if and only is $v$ is a leaf; if $ C(T)$ is a singleton $\{v\}$, $\Gamma(v,T)$ is the whole vertex set $V$; if $ C(T)$ is an edge $vv'$, $\Gamma(v,T)$ and $\Gamma(v',T)$ are the components of $(V, E\setminus\{C(T)\}$.

For a vertex $x$, denote $( x^0 \dots x^n )$ the unique minimal path in $ T $ connecting $x$ to the center: $x^0\in C(T)$, $x^n=x$; here $n$ is the distance $d(x):=n$ from $C(T)$. It is also convenient to consider the nested sequence $ \Gamma(x^i,T) $, and the vector $\gamma(x,T):=(\gamma_0,\dots,\gamma_n)\in\mathbb{N}^{n+1}$ whose $i$-th entry is the cardinality $\gamma_i:=|\Gamma (x^i,T)|$ of each of these sets. Note that, since the center of a tree is automorphism invariant, any automorphism of $T$ satisfy $\gamma(f(x),T)=\gamma(x,T)$. Among all leaves, consider those with minimum $\gamma(x,T)$ in the lexicographic order (with leading coefficient $\gamma_0$ ); we may shortly call them minimal leaves. For instance, the three leaves of the tree $E_7$ have labels and $(3,2,1)$, $(4,1)$, $(4,3,1)$ in increasing lexicographic order.

1. $\mathrm{diam}(T_x)=\mathrm{diam}(T)$. This means that $T$ and $T_x$ share a maximum length path, so they also have the same center. Thus, the entries of $\gamma(x',T_x)$ are simply $\gamma_i(x',T_x) = \gamma_i(x,T) -1$ for $i=0,\dots,n-1$. As a consequence, any automorphism $f$ of $T_x$ fixes the whole path connecting $x'$ to $C(T_x)=C(T)$ (this follows by induction on $i$, arguing on the cardinality of the connected components $\Gamma (x^i,T_x)$: now $\Gamma (x^0,T_x)$ has strictly minimum cardinality among the components of $(V_x, E_x\setminus \{C(T)\})$, so $f(\Gamma (x^0,T_x))=\Gamma(x^0,T_x)$ and $x^0$ is fixed; then $x^1$ is fixed because $\Gamma(x^1,T_x)$ has strictly minimum cardinality among the components of the sons of $x^0$ in $\Gamma (x^0,T_x)$, and so on ). Therefore, $f$ extends to an automorphism of $T$ that fixes $x$. Clearly, this is not the case if $T$ is a minimal element of $\mathcal{AFT}$.

either (2a) $T$ possesses a nontrivial automorphism (which is not the case if $T\in \mathcal{AFT}$), or (2b) $\mathrm{diam}(T)$ is an odd number $2n+1$, and the minimal path $(x^0,\dots, x^{n})$ connecting $x$ to $C(T)$ coincides with the whole $\Gamma(x^0, T)$. This is nice and not hard to prove; I'd add the details at your request.

Then, (by now on being in case 2b), since $(x^0,\dots, x^{n-1})$ is part of a maximum length path in $T_x$, and $ \mathrm{diam}(T_x)=2n $ is even, the center of $T_x$ is a single vertex, namely the other endpoint $y^0$ of $C(T):=\{x^0,y^0\}$.

  If moreover $T$ is a minimal element in $\mathcal{AFT}$, and $f_x$ denote the unique nontrivial automorphism of $T_x$, we know that $f_x(y_0)=y_0$ (it's the center of $T_x$), while $y:=f_x(x^{n-1})\neq x^{n-1} $ (otherwise $f_x$ would extend to $T$). Therefore, $ (y^0, f_x(x^0),f_x(x^1),\dots,f_x(x^{n-1}))$ is the $n$-edges path connecting $y$ to $C(T)$, and since the $x^i$ for $i\ge0$ are not branching points, this path has no branching points too, with the possible exception of $y^0$. Actually, $y^0$ must be a branching point, otherwise the path $\xi:=(x^n,x^{n-1},\dots,x^0,y^0,y^1,\dots y^n)$, which has maximal length $2n+1$ in $T$, would have no branching point at all, and therefore would be $T$ itself, what however is impossible because $T$ has no nontrivial automorphism.

Let me start with some general notions, that I believe are known, for a tree $T=(V,E)$ with finite, nonempty vertex set $V$ and edge set $E$. I will assume that $T$ is a minimal element of $\mathcal{AFT}$ only in the end.

For a path of length $n$ (number of edges) , $ (v_0 \dots v_n) $ in $ T $, let's define the centre of the path as the set $ \big\{v _ {\lfloor\frac{n }{2}\rfloor},v _ {\lceil \frac{n }{2}\\rceil } \big\}$, consisting of one or two vertices (thus, either the middle vertex, if $n$ is even, or the middle edge, if $n$ is odd). Given two paths, there is a third path including the centres of both, and one endpoint of each. As a consequence, all paths of maximum length in a tree share the same centre, that we can therefore refer to as centre of the tree, $C(T):=\{v,v'\}$ (so this notation allows that $C(T)=\{v\}=\{v'\}$, a singleton, precisely whenever the diameter of $T$ is an even number, as remarked).

The centre determines a natural genealogy order in $T$; in particular, we can attach to any vertex $v$ its progeny, $\Gamma(v,T)$, the set of all vertices $x$ such that the minimal path from $x$ to the centre passes by $v$. Thus, e.g. this reduces to $\{v\}$ if and only is $v$ is a leaf; if $ C(T)$ is a singleton $\{v\}$, $\Gamma(v,T)$ is the whole vertex set $V$; if $ C(T)$ is an edge $vv'$, $\Gamma(v,T)$ and $\Gamma(v',T)$ are the components of $(V, E\setminus\{C(T)\}$.

For a vertex $x$, denote $( x^0 \dots x^n )$ the unique minimal path in $ T $ connecting $x$ to the center: $x^0\in C(T)$, $x^n=x$; here $n$ is the path distance from $C(T)$. It is also convenient to consider the nested sequence $ \Gamma(x^i,T) $, and the vector $\gamma(x,T):=(\gamma_0,\dots,\gamma_n)\in\mathbb{N}^{n+1}$ whose $i$-th entry is the cardinality $\gamma_i:=|\Gamma (x^i,T)|$ of each of these sets. Note that, since the center of a tree is automorphism-invariant, any automorphism of $T$ satisfy $\gamma(f(x),T)=\gamma(x,T)$. Among all leaves, consider those with minimum $\gamma(x,T)$ in the lexicographic order (with leading coefficient $\gamma_0$ ); we may shortly call them minimal leaves. For instance, the three leaves of the tree $E_7$ have labels   $(3,2,1)$, $(4,1)$, and $(4,3,1)$, in increasing lexicographic order.

1. $\mathrm{diam}(T_x)=\mathrm{diam}(T)$. This means that $T$ and $T_x$ share a maximum length path, so they also have the same center. Thus, for any $v\in V_x$ we have $\Gamma(v,T_x)=\Gamma(v,T)\setminus\{x\}$, and in particular the entries of $\gamma(x',T_x)$ are simply $\gamma_i(x',T_x) = \gamma_i(x,T) -1$ for $i=0,\dots,n-1$. As a consequence, any automorphism $f$ of $T_x$ fixes the whole path connecting $x'$ to $C(T_x)=C(T)$ (this follows by induction on $i$, arguing on the cardinality of the connected components $\Gamma (x^i,T_x)$: now $\Gamma (x^0,T_x)$ has strictly minimum cardinality among the components of $(V_x, E_x\setminus \{C(T)\})$, so $f(\Gamma (x^0,T_x))=\Gamma(x^0,T_x)$ and $x^0$ is fixed; then $x^1$ is fixed because $\Gamma(x^1,T_x)$ has strictly minimum cardinality among the components of the sons of $x^0$ in $\Gamma (x^0,T_x)$, and so on ). Therefore, $f$ extends to an automorphism of $T$ that fixes $x$. Clearly, this is not the case if $T$ is a minimal element of $\mathcal{AFT}$.

either (2a) $T$ possesses a nontrivial automorphism (which is not the case if $T\in \mathcal{AFT}$), or (2b) no vertex of the minimal path $(x^0,\dots, x^{n})$ connecting $x$ to $C(T)$ is a branching point; moreover, if $T$ is assumed to be a minimal element of $\mathcal{AFT}$, $\mathrm{diam}(T)$ is an odd number $2n+1$. This is nice and not hard to prove; I'd add the details at your request.

Now, assume $T$ is a minimal element in $\mathcal{AFT}$; so we are in case in case 2b. Since $(x^0,\dots, x^{n-1})$ is part of a maximum length path in $T_x$, and $ \mathrm{diam}(T_x)=2n $ is even, the center of $T_x$ is a single vertex, namely the other endpoint $y^0$ of $C(T):=\{x^0,y^0\}$. If $f_x$ denote the unique nontrivial automorphism of $T_x$, we know that $f_x(y_0)=y_0$ (it's the center of $T_x$), while $y:=f_x(x^{n-1})\neq x^{n-1} $ (otherwise $f_x$ would extend to $T$). Therefore, $ (y^0, f_x(x^0),f_x(x^1),\dots,f_x(x^{n-1}))$ is the $n$-edges path connecting $y$ to $C(T)$, and since the $x^i$ for $i\ge0$ are not branching points, this path has no branching points too, with the possible exception of $y^0$. Actually, $y^0$ must be a branching point, otherwise the path $\xi:=(x^n,x^{n-1},\dots,x^0,y^0,y^1,\dots y^n)$, which has maximal length $2n+1$ in $T$, would have no branching point at all, and therefore would be $T$ itself, what however is impossible because $T$ has no nontrivial automorphism.