Let $G=(V,E)$ be an infinite simple, undirected graph with $\chi(G) \geq \aleph\_0$. Is there a minor $M$ of $G$ such that


1. $M\not\cong G$, and
2. $\chi(M)=\chi(G)$


?



Minors of graphs with infinite chromatic number
For every such $G$ there is an $M$ satisfies your requirement.


It is enough to show that for every graph $G$ with infinite chromatic number has two minors $G\_{0}$ and $G\_{1}$ such that 


$(i)$ $G\_{0}$ has no isolated vertex and $G\_{1}$ has exact one isolated vertex, and


$(ii)$ $\chi(G\_{0}) = \chi(G\_{1}) = \chi(G)$.


Proof:


Case 1: If $G$ has at least one isolated vertices, then remove all of them to get $G\_{0}$ and remove all but one of them to get $G\_{1}$.


Case 2: If $G$ has no isolated vertices, choose one of its vertices, $\nu\_{0}$, and remove all of edges which adjacent to $\nu\_{0}$ then remove all of its isolated vertices (if there are any) other than $\nu\_{0}$. We get our $G\_{1}$. 
It is easy to see that $ \chi(G\_{1}) = \chi(G)$. Since if there is a coloring function $c$ maps $V(G\_{1})$ to $\kappa \geq \omega$, then let $C$ be the following map from $V(G)$ to $\kappa$:


$(a)$ $C(\nu) = c(\nu) + 2$ if $\nu$ is neither $\nu\_{0}$ nor a leaf connected to $\nu\_{0}$,


$(b)$ $C(\nu) = 1$ if $\nu$ is a leaf connected to $\nu\_{0}$,


$(c)$ $C(\nu\_{0}) = 0$.


This conclude the proof.


