= Solution
Put $A=A_m$ and let $T$ be the <rooted tree> of finite words over $\{1,\ldots,m\}$. View $a\in A$ as a rooted <automorphism>: it permutes the first letter by $a$ and fixes the suffix. Denote this rooted copy by $A$ again.
For $a\in A$, define a directed tree <automorphism> $b(a)$ by putting local <permutation> $a$ at every vertex
$$
2,\quad12,\quad112,\quad1112,\quad\ldots
$$
and identity local <permutation> at every other vertex. These marked vertices are pairwise incomparable, so the prescribed actions are on disjoint rooted subtrees and define a unique <automorphism>. Equivalently its root <permutation> is trivial and its sections are
$$
b(a)=\bigl(b(a),a,1,\ldots,1\bigr).
$$
The map $a\mapsto b(a)$ is an <injective> <homomorphism>: it is the same copy of $a$ acting on each of those disjoint subtrees. Let $B=b(A)$ and define
$$
H=\langle A,B\rangle\le\operatorname{Aut}(T).
$$
Both $A$ and $B$ are finite, so $H$ is finitely generated. For $m\ge5$, $A_m$ is a nonabelian simple <group> and hence perfect. Thus $A,B\le H'$, and because they generate $H$, \b[$H$ is perfect].
Every section of either defining generator belongs to $H$, so sections of every element of $H$ belong to $H$. Its root <permutation> belongs to $A$. For a tree <automorphism> $g$, write $g(iw)=\sigma(i)g_i(w)$, where $g_i$ is its section at input letter $i$. In the semidirect-product convention of part (a), define
$$
\theta(g)=\bigl(g_{\sigma^{-1}(1)},\ldots,g_{\sigma^{-1}(m)};\sigma\bigr).
$$
Composition of tree automorphisms gives exactly the wreath-product multiplication: this is why $\theta$ is a <homomorphism>. It is <injective> because an <automorphism> is determined by its first-level <permutation> and all its first-level sections. On the generators,
$$
\theta(a)=(1,\ldots,1;a),\qquad
\theta(b(a))=(b(a),a,1,\ldots,1;1).
$$
It remains to prove surjectivity; merely writing these recursions would not establish it. Choose $\sigma=(2\,3\,4)\in A$, which fixes $1$ and sends $2$ to $3$. For $a,c\in A$, the two base tuples $\theta(b(a))$ and $\theta(\sigma b(c)\sigma^{-1})$ overlap only in their first coordinate. Their <commutator> is therefore
$$
\bigl([b(a),b(c)],1,\ldots,1;1\bigr)
=\bigl(b([a,c]),1,\ldots,1;1\bigr).
$$
Since $A$ is perfect, products of these commutators give every element of the copy of $B$ localized in coordinate $1$. Canceling this first coordinate from $\theta(b(a))$ then gives $(1,a,1,\ldots,1;1)$, the rooted copy of $A$ localized in coordinate $2$. Conjugating by the transitive rooted <group> $A_m$ supplies both localized $A$ and localized $B$ in every coordinate. Since $H=\langle A,B\rangle$, the image contains the full base $H^m$. It also contains the top $A_m$. Therefore it is the entire <wreath product> and
$$
\boxed{\theta:H\overset\sim\longrightarrow H\operatorname{wr}A_m.}
$$
Finally $H$ is infinite: it contains the nontrivial rooted $A_m$, and a finite order $h=|H|$ would satisfy $h=h^m|A_m|$, which is impossible for $m\ge5$.
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