Let $G \subseteq \mathbb F_p^*$ be a subgroup. Then $G$ is called almost trivial if $G \cap (2-G)$ consists of the element 1.
Then I am wondering how big $G$ can be in terms of $p$. If $G$ is a random set containing 1 then according to the birthday paradox one expects that $G$ and $2-G$ have nontrivial intersection as soon as $\#G \gg \sqrt p$.
I am wondering if maybe one could prove that all almost trivial subgroups of $\mathbb
F_p^*$ have size at most $\tilde O(\sqrt p)$.
Or if a $\sqrt p$ up to log factors bound is out of reach, I was wondering if there is $1/2 < \epsilon < 1$ and some constant $c$ such that for all almost trivial groups we have $\#G < cp^\epsilon$.
I did some numeric experimentation by finding the largest almost trivial subgroup of $\mathbb F_p^*$ for all p < 80000. And the largest ratio that I could find for $\frac {\#G}{\sqrt p}$ was $\frac {884}{\sqrt {41549}} = 4.3368\ldots$ because $\mathbb F_{41549}$ contains an almost trivial subgroup of order 884, so a square root (possibly up to log factors) bound seams reasonable.
P.S. The notion of being almost trivial is based on the notion of almost rational in the paper Almost rational torsion points on semistable elliptic curves by
Frank Calegari.
Best Answer
If we let $S$ be the set of characters of $\mathbb F_p^\times$ trivial on $G$ then $$\sum_{\chi \in S} \chi(g) = \begin{cases} \frac{p-1}{|G|} & g\in G \\ 0 & g\notin G \end{cases}$$
so $$\sum_{\chi_1,\chi_2\in S} \sum_{ g \in \mathbb F_p \setminus \{0,1,2\} } \chi_1(g) \chi_2(2-g) = 0$$ if $G$ is almost trivial.
Now if $\chi_1 =\chi_2=1$ then $\sum_{ g \in \mathbb F_p \setminus \{0,1,2\} } \chi_1(g) \chi_2(2-g)=p-3$ and otherwise, by the bound for Jacobi sums, $\left| \sum_{ g \in \mathbb F_p \setminus \{0,1,2\} } \chi_1(g) \chi_2(2-g) \right| \leq \sqrt{p}+1 $.
Thus, if $G$ is almost trivial then $p-3 \leq ( ((p-1)/|G|)^2-1) (\sqrt{p}+1)$, or $p \leq c^2 p^{5/2} / |G|^2$ for a constant $c$, meaning $|G| \leq c p^{3/4}$, answering the weaker form of your question.