Since we have only 3 points , say $A , B, C$ then $A$ has to be the circumcenter of the $\bigtriangleup ABC$ which is not possible

Let's say that our points are $A_i$ with $i \in \{1,2,3,4\}$. Then each $A_i$ is the circumcenter of $\bigtriangleup A_{i+1}A_{i+2}A_{i+3}$ then it's easy to see that all edges $A_iA_j$ have to be equal $\Rightarrow$ $\bigtriangleup A_1A_2A_3$ and $\bigtriangleup A_1A_2A_4$ are equilateral hence $\bigtriangleup A_3A_2A_4$ can't be equilateral since $\angle A_3A_2A_4 = 2 \cdot 60^{\circ} = 120^{\circ}$

We will say that $i \rightarrow jkl$ if $A_i$ is the circumcenter of $\bigtriangleup A_jA_kA_l$ $4 \rightarrow 123$ and $5 \rightarrow 234$. Note that $A_1$ can not be the circumcenter of the $\bigtriangleup A_2A_3A_4$ from case $n=3$ Case 1 : $1 \rightarrow 245$ or $1 \rightarrow 345$ WLOG $1 \rightarrow 245$ since the other one can be solved by the same arguments from symmetry Note that from the case when $n = 3$ $\Rightarrow$ we can't have $3 \rightarrow 124$ hence $3 \rightarrow 145$ or $3 \rightarrow 125$ The first one can not be true , because we would get that $\bigtriangleup A_1A_2A_3$ , $\bigtriangleup A_1A_2A_4$ and $\bigtriangleup A_1A_2A_5$ are all equilateral which is false. ( a diagram makes this a lot clearer). By similar arguments the latter case can't hold because $2$ can't be a circumenter of any triangle. Case 2 : $1 \rightarrow 253$ Then $A_1 , A_4$ and $A_5$ are collinear on the perependicular bisector of $A_2A_3$ and $A_1A_5 = A_1A_2$ , $A_4A_2 = A_1A_4$ and $A_4A_5 = A_2A_5$ . Now, a straightforward cartesian coordinates bash does the job with $A_2 = (-1,0)$ and $A_3 = (1,0)$ using the distance formula for the 3 relations from above.

I will add it later since there are more cases and it's a bit messy

n=6 is possible