If $G=(V,E)$ is a finite, simple, undirected graph, the Hadwiger number $\eta(G)$ is defined to be the size of the largest complete minor of $G$.
Is there a finite graph $G=(V,E)$ with the following property?
Whenever $v,w\in V$ have distance $2$, the Hadwiger number of the graph obtained by collapsing $v,w$ is strictly larger than $\eta(G)$.
Well, if you collapse all vertices of distance 2 in a connected graph, then the resulting graph will either be a single vertex (if the graph is not bipartite) or an edge.
But there are graphs where you can identify some of the vertices of distance 2 from each other to increase $\eta(G)$.
Indeed,
First let $G$ be a $K+1$-clique where $K$ is of the form e.g. $K=d(d-1)$ for some integer $d \geq 3$.
Now subdivide each edge in $G$ to a path with say 5 vertices and call the resulting graph $G'$. Then every vertex of degree greater than 2 in $G'$ has degree $K$, and $\eta(G)$ is still $K+1$.
Now replace every vertex $v$ in $G'$ of degree $K$ with a complete $d$-ary tree $T_v$ with $K=d(d-1)$ leaves and write the set of $K$ leafs in $T_v$ as $\{u_{vv'}; v'$ another vertex in $G'$ of degree $K\}$--for every two vertices $v,v'$ of degree $K$ in $G'$, there is now a path from $u_{vv'}$ of $T_v$ to $u_{v'v}$ of $T_{v'}$ (every vertex and edge in the interior of this past is the same in $H$ as it was in $G'$). Call the resulting graph $H$.
Then $\eta(H) \le d < K < \eta(G)$. However, identifying in $H$ all vertices of distance 2 in each of the $T_v$s results in a graph that is close enough to $G'$ and as rthe same Hadwiger number as $G'$. Indeed this collapses each $T_v$ to an edge where one endpoint has degree $K$ (and the other endpoint is incident to no more edges). In fact this resulting graph has $G'$ as a subgraph--indeed every vertex of degree $K$ in $G'$ is incident to another edge where the endpoint is an isolated vertex.
