Consider an $n$-vertex multigraph. Every edge is coloured by one of $n$ colours and the edges of each colour form a cycle of odd length. Prove that there is a cycle of odd length whose edges are all of a different colour.
If we were to consider a normal complete graph, then for cases where the number of edges is at least $n$ we could prove inductively that the graph needs to contain a different-colour cycle of length $3$. Otherwise, if the number of edges is $\le n$, we would have $\frac{n(n-1)}{2}\lt n$, therefore $n=2$. In the case $n=2$ the condition cannot be met. Thus, it holds.
I do not know how and whether this is relevant to the problem of the general multigraph. Any help will be appreciated.
If you take one edge of each cycle, you end up with a graph of $n$ edges on $n$ vertices. So you must have a cycle, which one will have all edges of different colors.
To find an odd length cycle, choose an edge for each colored cycles one by one in the following. For each cycle, if choosing it keeps the graph a forest, then choose it. At one point, for one cycle, every edge added will form a cycle. Represent your forest as a bipartite graph. You cycle must have at least one edge contained in one side of the bipartite graph. As this edge must form a cycle, it must form an odd length one.