Given $G$ a regular graph on $n$ vertices, denote $\alpha(G)>1$ to be independence number.

Denote $\Gamma(G)$ to be collection of possible subset of independent vertices in $G$ of cardinality $\alpha(G)-1$.

To each $\gamma\in\Gamma(G)$, assign number $N(\gamma)$ reflecting number of ways $\gamma$ could be extended by an additional vertex so that augmented subset remains independent (attains cardinality number $\alpha(G)$).

Denote $N(G)=\max_{\gamma\in\Gamma(G)}N(\gamma)$.

Denote $M(G)$ to be maximum number of disjoint independent sets of $G$ that attain cardinality $\alpha(G)$ (that is each subset in $M(G)$ should be disjoint with cardinality $\alpha(G)$).

Easy to observe that $M(G)\leq\frac{|V|}{\alpha(G)}$.

Given fixed real $r>3$ (example $3.00002$), is there a graph (family) such that $$M(G)>|V|^{\frac{r-1}{f(r)}}>|V|^{\frac{1}{f(r)}}> \max(N(G),\alpha(G))$$ where $|V|$ is vertex number with some function $f(r)\geq r$?

small EXAMPLEwas fine, I was not confused at that stage. $\endgroup$onlymy accidental conclusion was wrong (a result of a mistakenthinkingat that moment--my concentration gave up when I mixed the general definition and the peculiarities of my construction). Another equivalent formulation: $\ m:=M(G)\ $ is the largest integer such that there exists $\ W\subseteq V\ $ such that $\ |W|=\alpha(G)\cdot m\ $ and W is a union of $\ m\ $ maximal independent sets (i.e. od independent sets $\ J\subseteq V\ $ such that $\ |J|=\alpha(G)$). $\endgroup$4more comments