Let $ABCDE$ be a convex pentagon of perimeter $\mathcal{P}$. Consider $F,G,H,I,J$ the mid points of $\overline{AB}$, $\overline{BC}$, etc. Denote $\mathcal{Q}$ the perimeter of the pentagon $FGHIJ$.
If $ABCDE$ is regular, with high school geometry tools we can easily show that $\frac{\mathcal{Q}}{\mathcal{P}} = \frac{1+\sqrt{5}}{4} = \sin (3\pi/10)$.
I am interested to show that for every convex pentagon (not just regular) we have $\frac{\mathcal{Q}}{\mathcal{P}} \leq \frac{1+\sqrt{5}}{4}$. I do not have a proof (yet) but computer simulations seem to indicate that the result is true.
Like many problems of the sort, regularity offers a maximum (for instance isoperimetric inequalities https://en.wikipedia.org/wiki/Isoperimetric_inequality).
If we cannot get $\frac{1+\sqrt{5}}{4}$, I would still be interested in a weaker result, for instance:
There exists $c<1$ such that for every convex pentagon, we have $\frac{\mathcal{Q}}{\mathcal{P}} \leq c$.
COMMENT:
If we continue construction using mid points, we can see that all pentagons are similar in regular and irregular pentagons. Let's denote the perimeters as $P_1$, $Q_1$, $P_2$, $Q_2$ . . etc. We have:
$\frac{2Q_1}{P_1}=\frac{2Q_2}{P_2}=\frac{1+\sqrt 5}2$
This is golden ratio in Fibonacci sequence. Considering the fact that an irregular polygon can be assumed as a transformed regular polygon ;this is due to the fact that the sum of angle is always $(n-2)\pi$ , we may conclude that the ratio of perimeters obey similar rule.