Warning: too chatty answer. Sorry about that.
TL;DR If all the corner points are given, then everything you need is in the first chapter of any computational geometry book.
For $1 \le i \le 6,$ let the corner points be $p_i = (x_i, y_i).$ Let the sides of the polygon be $$s_1 = (p_1, p_2), s_2 = (p_2, p_3), \ldots, s_5 = (p_5, p_6), s_6 = (p_6, p_1)$$
Of course, the length of each side $s = (p_i, p_j)$ is given by $\| p_i - p_j \|.$
Since we are given all the $p_i$'s, we can form linear equations for the $6$ lines passing through each side. Each line takes the forms $$ l_i : y = m_i x + b_i $$
where $m_i$ is the slope, and $b_i$ is the intercept. Of course, solving each pair of lines would give one of our points back. For example, solving $l_1$ and $l_6$ would yield $p_1.$
Now, given that the new polygon is $d$ meters away (sorry I used $d$ instead of $x$ to avoid notation conflict), our goal should be
Find the $6$ new line equations $l'_i: y = m'_i x + b'_i$
Solve each pair of $l'_1, l'_2$ etc to find all corner points $p'_i.$
From $p'_i,$ compute the new side lengths, and hence the scaling factor for each side.
Step 2, 3 are obvious. So I'll focus on step 1. To do so, let's look on the new system of line equations $$ l'_i : y = m'_i x + b'_i $$ corresponding to the lines passing through the sides of the scaled away polygon. To find $m'_i$ and $b'_i,$ we first notice that the slope remains the same, but the intercept changes. In other words, $$ m'_i = m_i $$
Now we're left with finding $b'_i,$ and the problem is solved.
To do so, take each pair of line equations and do the following:
I'll only pick, for example, $l_1: y = m x + b_1$ and $l'_1: y = mx + b'_1$ (we already agreed that $m'_1 = m'_2 = m.$) Remember, we know $l_1,$ and we know a point $p_1$ on $l_1.$ We have to figure out $b'_1.$ The perpendicular distance from $l_1$ to $l'_1$ is $d.$ i.e., the distance from $p_1$ to $l'_1$ is $d.$
Out of laziness, I will leave the rest as an exercise. Given $p_1,$ and $l_1,$ figure out the unit vector in the direction of $l_1.$ Then flip it (sign change) to get the unit vector in the normal (perpindicual) direction on $l_1.$ Translate $p_1$ in the direction of the normal $d$ meters away to get $p'_1.$ We got our first point! Solve $l'_1$ using $p'_1$ to find $b'_1.$ Repeat for all pairs of $l_i, l'_i.$ And continue the steps 1, 2, 3 above.