Consider the action of multiplying by $(n-1) \pmod n$, in the group of units $U(n)$ when $n$ is an even number. This action is a map $f$ defined on $U(n)$ such that for each $x \in U(n), f(x) = (n-1)x \pmod n$.
Remember that $U(n)$ is formed by the predecessors of n that are relatively prime to $n$. The order of the element $n-1 \in U(n)$, when $n$ is even, is always $2$. Therefore, the orbits of that action will always have two elements. Moreover, the elements in these orbits will be additive inverses of the group of integers $\pmod n$.
For instance, in the group of units of multiplication $\pmod {10}$, denoted by $U(10)$, the orbits of the action of multiplying by $9 ~(=10-1)$ are $\{1, 9 \}$ and $\{3, 7 \}$.
Here, we see that there is at least one orbit, $\{3, 7 \}$, formed exclusively by prime numbers.
Therefore, to solve Goldbach's conjecture, all we have to do is explain why there is always at least one orbit formed by prime numbers in the action of $(n-1)$ in $U(n)$, where $n$ is even. Correct?