Something seems very odd here. The starting material is C14H11O2N, and the "rearrangement" product is C14H11ON. In other words, this reaction represents the loss of O, and is therefore a reduction. You do not have any reducing agents listed, just sulfuric acid: did you leave anything out?
Also, I infer you add nitrosobenzene in the second step. This reaction is simple enough; ethoxide deprotonates the N-oxide, the double bond on the indole adds at the 3-position to N=O, and you get a bond. Proton shuffle to protonate the resulting N-O- compound, then another equiv. of EtO- takes off the 3-indole proton and eliminates OH- in an E2 reaction between the 3-indole carbon and the nitrosobenzene nitrogen, and you get a double bond between the two atoms. The net result is the formation of a double bond and water as a byproduct. Very similar to the condensation of an arylamine and a nitrosobenzene to form an azobenzene and water.