Minimal number of vertices in a graph with special Hadwiger partitions

Question

Let $G=(V,E)$ be a simple, undirected graph.

We call a partition ${\cal P}$ of a non-empty subset of $V$ a Hadwiger partition if

1. every block (member of ${\cal P}$) is non-empty and connected, and
2. if $x, y \in {\cal P}$ are distinct blocks then there are $v\in x$ and $w \in y$ such that $\{v,w\} \in E$.

The Hadwiger number $\eta(G)$ is the maximum number of blocks a Hadwiger partition can have.

Let $n=\eta(G)$ and suppose that

1. every Hadwiger partition of $G$ into $n$ blocks contains at least one block consisting of 1 vertex only ("singleton block"), and
2. for all $v\in V$ we have $\eta(G\setminus\{v\}) = n$.

Does this imply that $|V|\geq 2n$?

MOTIVATION: In the answer to this question, @bof gave excellent and short examples for graphs fulfilling conditions 1 and 2 above, each having exactly $2n$ vertices. It's not hard to find other examples with at least this number of vertices. So now, it's open season for smaller examples -- if they exist!

Answer 1

Octahedron$ .$