= A finitely generated perfect group isomorphic to its alternating wreath product
For every $m\ge5$, let $A=A_m$ act by rooted permutations on the rooted tree of finite words over $m$ letters. Define $b(a)$ to have local permutation $a\in A$ at the pairwise incomparable vertices $1^r2$ for $r\ge0$, and identity elsewhere. Then $b$ is an embedding of $A$ and satisfies $b(a)=(b(a),a,1,\ldots,1)$ with trivial root permutation. The <group> $H=\langle A,b(A)\rangle$ is finitely generated and perfect because it is generated by two finite perfect subgroups.
The root permutation and sections define an injective <homomorphism> $H\to H\operatorname{wr}A_m$. To prove surjectivity, conjugate $b(c)$ by the rooted cycle $(2\,3\,4)$ and take its commutator with $b(a)$. In wreath coordinates this leaves only $b([a,c])$ in the first coordinate. Perfectness of $A_m$ therefore supplies the entire localized copy of $b(A)$ there. Canceling it from the recursion for $b(a)$ supplies a localized rooted $a$ in the second coordinate. Transitivity moves both kinds of generators to every coordinate, yielding the full base $H^m$ and hence the full wreath product. The group is infinite, since a nontrivial finite order $h$ could not satisfy $h=h^m|A_m|$.
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