EDIT: This question is different from the other in a sense that unlike it this one goes into specifics and is intended to solve the problem. In the previous post, the only answer was a hint. In this post, the answer is a proper solution. Is there a way to polynomialy reduce the problem of finding graph wheel $W_{n}$ to the problem of detecting Hamiltonian Cycle (or vice versa) ? The ultimate goal is to show that finding $W_n$ is NP-complete problem. So far I can show that the problem is in NP. To show that I claim that given an input of graph it is possible in polynomial time to check every vertex and to make sure that it has three edges(except the hub): one goes to the hub, one goes to the next vertex and one to the previous vertex. By checking every vertex in this fashion one can make sure the ticket indeed represents $W_{n}$.As far as converting from the Hamiltonian Cycle problem to the problem of fining $W_n$ goes I have encountered that $W_n$ has many Hamiltonian cycles inside it.It is easy to see if one imagines a wheel with  a hub and bunch of spokes. Every triangle created by the spokes forms a Hamiltonian cycle as well as any other closed loop formed inside a given $W_{n}$. The intriguing part is the fact that only one vertex is in the way and that vertex is the hub. Is the hub of the wheel main culprit in figuring out the conversion ?
Reducing from Hamiltonian Cycle problem to the Graph Wheel problem
graph theory;np complete
We will add one extra vertex $v$ to the graph $G$ and we make new graph $G'$, such that $v$ is connected to the all other vertices of $G$. $G$ has a Hamiltonian cycle if and only if $G'$ has a $W_{n+1}$. It is easy to check that if $G$ has a Hamiltonian cycle then $G'$ has a $W_{n+1}$ wheel (just set $v$ as a center). On the other hand, if $G'$ has a $W_{n+1}$ then there are two possibility:$v$ is the center of $W_{n+1} \rightarrow G $ has a Hamiltonian cycle.Another vertex $u$ is the center of $W_{n+1}$ in $G'$. So both $deg(u) = deg(v) = n$. Then we can change the labeling of this two vertices (actually they are equivalence under isomorphic), now we have again first possibility.P.S1: By $W_n$ I mean the wheel with $n$ vertex.P.S2: In this proof we say if we fix $k=n+1$ (size of the artificial graph), then the problem is NP-Complete in this restricted version, So it's also NP-Complete in the case $k$ is as input parameter.