TITLE: Can a graph be reconstructed from its cycle lengths?
QUESTION [11 upvotes]: All graphs discussed are finite and simple. The cycle sequence of a graph $G$, denoted $C(G)$, is the nondecreasing sequence of the lengths of all of the cycles in $G$, where cycles are distinguished by the vertices they contain, not by the edges they contain. 
For example, $C(K_{3,2})=4,4,4$ and $C(K_4)=3,3,3,3,4$.
Two graphs are isoparic if they have the same number of vertices and the same number of edges. 
Main question: If $G$ and $H$ are 2-connected nonisoparic graphs, can $C(G)=C(H)$?
The 2-connected condition is so we can't just make a bunch of edge-disjoint cycles that share a vertex. The nonisoparic condition is so we can ignore situations like the following:

These graphs are not isomorphic but are isoparic. Both graphs have the cycle sequence $3,3,4,5,5,6$ and can be viewed as just a square surrounded by two triangles. Perhaps there's a better way to ignore this trick besides the nonisoparic condition.
I'm interested more generally in finding out exactly what the cycle sequence can tell us. When is a cycle sequence realizable by a 2-connected graph? Is such a realization ever unique? I've looked at a couple dozen graphs on fewer than seven vertices and the only duplicate cycle sequences have been for the graphs shown above. 
Thank you.

REPLY [10 votes]: Second Answer
I'm adding this as another separate answer, rather than editing the first "answer" because otherwise anyone coming late to this discussion will end up doubly confused.
So let's try again, and say that the answer to your question is still "Yes". 
If you type the following into Sage 
g1 = Graph("G?rFf_")
g2 = Graph("H??EDz{")

and then show them as before, we get


then I think that they each have exactly 11 4-blobs and 4 6-blobs (using "blob" rather than overloading the word cycle) but one has 8 vertices and 12 edges and the other has 9 vertices and 13 edges.
Here's a list of the blobs for the first graph (preceded by the size)
4 5 4 1 0
4 6 4 1 0
4 6 5 1 0
4 7 4 1 0
4 7 5 1 0
4 7 6 1 0
4 7 6 2 0
4 7 6 2 1
4 7 6 3 0
4 7 6 3 1
4 7 6 3 2
6 7 6 4 2 1 0
6 7 6 4 3 1 0
6 7 6 5 2 1 0
6 7 6 5 3 1 0

and here's the ones for the second graph
4 8 6 1 0
4 8 7 2 0
4 8 7 3 0
4 8 7 3 2
4 8 7 4 0
4 8 7 4 2
4 8 7 4 3
4 8 7 5 0
4 8 7 5 2
4 8 7 5 3
4 8 7 5 4
6 8 7 6 2 1 0
6 8 7 6 3 1 0
6 8 7 6 4 1 0
6 8 7 6 5 1 0