### This conjecture is false.

For odd and , let be the graph obtained by deleting an edge, say , from the Petersen graph. Define to be the graph obtained by joining vertex and vertex with an edge (with subscripts reduced modulo ). For each , the set is an edge cut. Hence, in any 2-factor of , either none of the edges of the form are contained in a cycle, or all of them are contained in the same cycle.

Case 1: If none of the edges described above are contained in a cycle of a 2-factor of , then this 2-factor contains a 2-factor of for each . These 2-factors are also 2-factors of the graphs , that is, each is a 2-factor of the Petersen graph. Each 2-factor of the Petersen graph consists of two cycles of 5 vertices, hence, any such 2-factor of contains cycles of odd length.

Case 2: See the comment below.