Suppose I have a strongly $k-regular$ graph $G$, of size $v$, where every vertex is $N>0$ $n-cycles$, for $at least$ one value of $n$ that divides $v$. Can we cut edges from $G$ in such a way that we're left with a graph of $v/n$ disconnected components, where each component is an $n-cycle$ for at least one value of $n$?
I've made some edits to the original questions to improve conditions and clarity
I use some non-standard terms here:
My answer will not address your theorem specifically, but instead I want to ask: when does there exist a $n$-uniformly cyclic vertex subdecomposition of a graph?
First we will define a vertex decomposition of a graph. This is a term that is in relation to the regular decomposition of graphs:
A decomposition $\mathfrak{G} := \{ (v_i \subset V,s_i \subset E) \}$ of a graph $G=(V,E)$ is a family of edge induced subgraphs such that $$ \bigcup^{n} s_i = E$$ and the $s_i$ are pairwise disjoint.
A vertex decomposition $\mathfrak{V} := \{ (v_i \subset V,s_i \subset E) \}$ of a graph $G=(V,E)$ is a family of vertex induced subgraphs such that $$ \bigcup^{n} v_i = V$$ and the $v_i$ are pairwise disjoint. (I c
A vertex subdecomposition $\mathfrak{S}$ of $\mathfrak{V}$ is a family of subgraphs of $\mathfrak{V}$ such that $$ \bigcup^{n}_{\mathfrak{S}} v_i = V $$.
We are looking for specific vertex subdecompositions of graphs. Specifically, we want to know when is every element in $\mathfrak{S}$ a cycle of degree $n$. We call this a $n$-uniformly cyclic vertex subdecomposition because the cycles are of uniform degree $n$.
An important consequence of the fact that the vertex decomposition is a family of vertex induced subgraphs that do not share vertices is that any vertex decomposition will be a disjoint graph, as we desire.
Your conjuncture is equivalent to the conjuncture that: There exist an $n$-uniformly cyclic vertex subdecomposition $\mathfrak{S}$ of a graph $G$ if $G$ is $k-regular$, $n|v$ (where $v = |V|$), and every vertex of the graph is in $N>0$ $n$-cycles.
We know that if there exists a vertex decomposition $\mathfrak{V}$ of $G$ where every $v_i$ has equal cardinality and if for every $g_i \in\mathfrak{V}$ there is an $n$-cycle with is a spanning subgraph $g_i$ then there exist an $n$-uniformly cyclic vertex subdecomposition $\mathfrak{S}$ of a graph $G$. Therefore it is sufficient to show that a given graph $G$ has a vertex decomposition $\mathfrak{V}$ of $G$ where every $v_i$ has equal cardinality and if for every $g_i \in\mathfrak{V}$ there is an $n$-cycle with is a spanning subgraph $g_i$.
The first condition is easily met if $n|v$ since we can construct a vertex decomposition where all $g_i$ have equal carnality by simply partitioning the vertices into $\frac{v}{n}$ partitions, where each one will have $n$ vertices. Further, there are $\binom{v}{n}$ such partitions.
Now we must figure out when one of those partitions has the property that every $g_i$ has a spanning subgraph which is an $n$-cycle. This is equivalent to the subgraph having a hamoltonian cycle since they all have degree n. This is also equivalent to the graph $G$ being $2$-factorable, and every $2$-factor having cardinality $n$. Since a graph can be $2$-factored if and only if it is a $K$-regular graph, where $K$ is even, then we know that the following condition must also be true: $G$ is $K$-regular, where $K$ is even.
So, we have a few necessary conditions, but no sufficient conditions:
If there exists $n$-uniformly cyclic vertex subdecomposition of $G$, then $n|v$, and $G$ is $K$-regular.
It is not true.
Take a non-bipartite cubic graph with no 3-cycles. Let $G$ be its line-graph. Then every vertex of $G$ lies on exactly two 3-cycles but there is no collection of disjoint 3-cycles covering every vertex.
