Fitting subgroup not is isomorph-free

Statement
The Fitting subgroup of a group need not be an isomorph-free subgroup.

Related facts

 * Solvable core not is isomorph-free
 * Fitting subgroup is normal-isomorph-free in finite
 * Solvable core is normal-isomorph-free in finite
 * Perfect core is homomorph-containing

Example of the symmetric group
Let $$G$$ be the symmetric group on the set $$\{ 1, 2, 3, 4 \}$$. The Fitting subgroup of this is a Klein-four group described as:

$$H := \{, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}$$.

$$H$$ is not isomorph-free in $$G$$: it is isomorphic to the group:

$$K := \{, (1,2), (3,4), (1,2)(3,4) \}$$.

Another generic example: the product of a Fitting-free group with a nilpotent group
Suppose $$H$$ is a nilpotent group and $$K$$ is a Fitting-free group, containing a subgroup $$L$$ isomorphic to $$H$$. Let $$G = H \times K$$. The Fitting subgroup of $$G$$ is $$H \times 1$$. However, this subgroup is isomorphic to $$1 \times L$$.

An example might be to take $$K$$ as any non-Abelian simple group, and $$H$$ as isomorphic to an Abelian subgroup $$L$$ of $$K$$. For instance, $$K$$ is the alternating group on five letters and $$H$$ is a cyclic group of order two.