The question should be clarified.

Let me assume the question is as follows (I couldn't think of another nontrivial interpretation):

Does there exist a free group $F\neq 1$ and a finite group $L$ acting on $F$ such that the semidirect product $F\rtimes L$ is perfect?

The answer is yes with $F=F_9$.

Indeed, start from $H$ the derived subgroup of index 2 in the signed permutation group $C_2\wr A_5$. So $H$ has order $2^4.60$ and is perfect.

Let $F_5$ be freely generated by $x_1,\dots,x_5$. Then $H$ naturally acts on $F_5$: the alternating group permutes the generators, and the normal abelian subgroup acts by changing the sign of generators (because we restrict to $H$, we consider this changing the sign of two coordinates). For instance the element $x_i\mapsto x_i^{-1}$ for $i\le 2$ and $x_i\mapsto x_i$ for $i\ge 3$ is such an element.

Then the semidirect product is not perfect, but its subgroup of index $2$, $F_9\rtimes H$ is perfect. Here $F_9$ consists of elements of even length in $F_5$.

The question should be clarified. (done)

Let me assume the question is as follows (I couldn't think of another nontrivial interpretation):

Does there exist a free group $F\neq 1$ and a finite group $L$ acting on $F$ such that the semidirect product $F\rtimes L$ is perfect?

The answer is yes with $F=F_9$.

Indeed, start from $H$ the derived subgroup of index 2 in the signed permutation group $C_2\wr A_5$. So $H$ has order $2^4.60$ and is perfect.

Let $F_5$ be freely generated by $x_1,\dots,x_5$. Then $H$ naturally acts on $F_5$: the alternating group permutes the generators, and the normal abelian subgroup acts by changing the sign of generators (because we restrict to $H$, we consider this changing the sign of two coordinates). For instance the element $x_i\mapsto x_i^{-1}$ for $i\le 2$ and $x_i\mapsto x_i$ for $i\ge 3$ is such an element.

Then the semidirect product is not perfect, but its subgroup of index $2$, $F_9\rtimes H$ is perfect. Here $F_9$ consists of elements of even length in $F_5$.