Rob, if you poke around on the internet you can find many sites that show diagrams for how the $\ce{N2}$ molecular orbitals are formed from two sets of nitrogen ($\ce{N1}$) atomic orbitals. Many of the diagrams are different one from another - that's a tip off that s-p mixing is not something that can always be predicted in advance. With that understanding, let's step through your diagram and questions.
It might be useful to take a look at the $\ce{N2}$ MO diagram here. It's just like your diagram. but they break down the steps and it might be a bit less confusing.
In any case we have a nitrogen atom with 5 valence electrons, 2 in the 2s atomic orbital and 1 each (total of 3) in the 3 degenerate p orbitals. If we move a second nitrogen atom into the vicinity, these atomic orbitals will hybridize in order to start the bond formation process.
$\ce{N2}$ is analogous to acetylene, they both have a triple bond that is made up of 1 sigma bond and 2 pi bonds. In acetylene - and nitrogen - the sigma bond is $\ce{sp}$ hybridized. We form this bond by
- Mixing the s and a p atomic orbitals (the p orbital whose lobe is linear with the s orbital [see the diagrams in the link I supplied above, they call it the $\ce{p_{z}}$ orbital]) on one of the nitrogens creates a bonding and antibonding $\ce{sp}$ orbital on that nitrogen atom.
- Same thing happens on the other nitrogen atom
- As the 2 nitrogen atoms begin to interact and form bonds, the 2 $\ce{sp}$ orbitals on each nitrogen interact (split further).
- The bonding $\ce{sp}$ orbital on each nitrogen interacts to form a lower energy and higher energy pair of bonding molecular orbitals (the $\ce{1\sigma _{g}}$ and $\ce{1\sigma _{u}}$ orbitals in your diagram - BTW, there should be a dashed line from the p orbitals to the $\ce{1\sigma _{g}}$ orbital - it is missing in your diagram, maybe that has led to some of your confusion).
- The antibonding $\ce{sp}$ orbital on each nitrogen also interact in the same manner to create 2 antibonding sigma molecular orbitals ($\ce{2\sigma _{g}}$ and $\ce{2\sigma _{u}}$)
Now for pi bonds,
- The remaining two p atomic orbitals on each nitrogen are orthogonal to the s and $\ce{p_{z}}$ orbital - they can't mix with those orbitals.
- These atomic p orbitals $\ce{p_{x}}$ and $\ce{p_{y}}$ can only mix with their like counterpart on the other nitrogen atom. This gives rise to a bonding and antibonding $\ce{p_{x}}$ pi molecular orbital and a bonding and antibonding $\ce{p_{y}}$ pi molecular orbital
With that background, let's now look at your questions
s-p mixing in N 2 causes the 1σ g orbital to decrease in energy,
the 2σ g orbital to increase in energy (to higher than the 1π u )
yet the 1σ u is left unchanged by the interaction. Why is this?
I'd say it a bit differently as I did above. Mixing of the 2 bonding $\ce{sp}$ orbitals, one on each nitrogen atom, causes them to split (one is raised and one is lowered). The same thing happens to the antibonding $\ce{sp}$ orbital on each nitrogen - one is raised, one is lowered.
Clearly they are the s and p orbitals but is this within the atom or
is the s orbital on one atom interacting with the p on the other?
Hopefully I answered this above. I tried to distinguish between atomic and molecular orbitals. If I haven't been clear, post back.
Some clarification as to why the 1σ g orbital to decreases in
energy, the 2σ g orbital to increases in energy would also be much
appreciated.
Hopefully what I wrote above clarified this as well, let me know if it didn't.
