Pronormality is not centralizer-closed

Statement
It is possible to have a group $$G$$ and a pronormal subgroup $$H$$ of $$G$$ such that the centralizer $$C_G(H)$$ is not pronormal.

Related facts

 * Pronormality is normalizer-closed: In fact, normalizer of pronormal implies abnormal
 * Pronormality is not commutator-closed
 * Automorph-conjugacy is centralizer-closed
 * Isomorph-conjugacy is not centralizer-closed

Example of the symmetric group
Suppose $$G$$ is the symmetric group on the set $$\{ 1,2,3,4 \}$$ and $$H$$ is a $$2$$-Sylow subgroup of $$G$$, given by:

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

Then, $$H$$ is a pronormal subgroup of $$G$$, because it is a Sylow subgroup and Sylow implies pronormal. On the other hand, we have:

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

This is not pronormal, because it is conjugate to the subgroup $$\{, (1,2)(3,4) \}$$ via the permutation $$(2,3)$$, but these are not conjugate in the subgroup they generate.

Let $$G$$ be the symmetric group on the set $$\{ 1,2,3,4 \}$$.