What happens when I connect the GND of a LDO in a Diode to the real GND?
Yes, if you raise the LDOs GND pin, the output will rise accordingly. This is because the LDO now see's a higher voltage between it's reference point and ground.  

All a fixed regulator is, is a standard regulator with the programming resistive divider inside the IC, rather than outside it.  



If you look at the above image (imagine the resistive divider is inside the IC), you can see if you raise the bottom of the Vref pin, the output voltage will have to rise in order to match the divider input of the comparator with the divider input. As Photon comments though, it's not so accurate though to use a diode (unless you use a more complex circuit which can compensate for temp variations, etc, but then you may as well buy an adjustable regulator)

In the LM7805 datasheet, there is an example of how to create and adjustable regulator by driving the reference input with an opamp:



And a slightly simpler version which uses a fixed divider and no opamp:



For this version, you can use the formula provided - if you use a 5V regulator and want 5.3V, then:

5V * (1 + (3k&Omega; / 50k&Omega;)) = 5.3V

So R2, = 3k&Omega;, and R1 = 50k&Omega; (if you want exact you need to take into account the error from the Iq*R2 bias current, which is also given in the formula - the above is just a basic "near enough" example)