Not really, no. Clusters, especially as determined by k-means, do not have "features". The k-means documentation indicates,
This iterative partitioning minimizes the sum, over all clusters, of the within-cluster sums of point-to-cluster-centroid distances. Rows of X correspond to points, columns correspond to variables.
If you were working with two variables (two columns) then a geometric interpretation would be that any given point P is considered to be associated with a centroid C if the circle of radius |PQ| around P does not (strictly) include any other centroid.
Likewise you can extend to three variables (three columns) and extend the geometric interpretation to be "sphere" instead of "circle".
You can see, then, that the cluster associated with C does not have any particular value or even value range for the variables. The boundary between clusters would be the intersection of hyperspheres around the centroids.
If you were using SVM, and you had only two clusters, then you could find a hyperplane that divided the two clusters -- but a hyperplane is not what one would typically think of as being a "feature".
