Let $T$ be a local minimum spanning tree that is not a global minimum.
Let $e_1,\ldots,e_m$ the edges of $T$ ordered by non-decreasing weight.
Let $T'$ be a minimum weight spanning tree that coincides with $T$ on the largest possible begin segment,
so we may assume $T'$ has edges $e_1,\ldots,e_{n-1},f_n,\ldots,f_m$ (again ordered by non-decreasing weight)
and $f_n\ne e_n$. Let $w$ be the weight function in our graph.
We claim that $w(f_n)<w(e_n)$: indeed, if $w(e_n)\leq w(f_n)$ then $e_1,\ldots,e_{n-1},e_n$ would
be a valid startup for Kruskal's algorithm and would lead to a minimum weight spanning tree that
coincides with $T$ on a larger initial segment.
$T+f_n$ contains exactly one cycle $C$. The edges of $C$ different from $f_n$ cannot all be in
$\{e_1,\ldots,e_{n-1}\}$ or we would have a cycle in $T'$. So we find at least one edge $f$ on $C$ that
is different from $e_1,\ldots,e_{n-1}$ and $f_n$. But then $w(f)\geq w(e_n)>w(f_n)$,
so $T+f_n-f$ is a spanning tree with a smaller total weight that is a neighbour of $T$ in the spanning tree graph. Contradiction.
(I left some small details for you to prove. Let me know if they cause trouble).
Note that this proves the slightly stronger statement, that each non-minimal tree has
a neighbour of strictly smaller total weight.
It just means that you have a set of $n/2$ spanning trees where no two trees in the set have an edge in common. So it is a property of pairs of trees, not a single tree. Thus it is trivially satisfied for the only connected graph on $n=2$ verices.
Best Answer
This is a construction that gives $6n^2-n$