Voltage doesn't matter without information about current.
Your power requirements
You have 64 LEDs there, but at same time you probably have 50% of them turned on.
Typical LED maximum current is between 10-20mA.
Your LEDs need 2.3-3.3V but your supply is 5V. You should reduce supply voltage to 4V (as Ignacio suggested in comment).
P = 4V * 15mA = 60mW
For 32 diodes:
P = 32 * 60mW = 1920mW = 1.92W
You will need 10W (thats at least 30x30cm size) or even much bigger solar panel to power it in building (no direct sun). Solar panel rated at 10W have maximum power of 10W, and thats rating for direct sun light.
So - technically - thats bad idea.
However you can try to extend time on battery with few small solar panels. They will look good in your art.
If you are really want to use solar power
You can use one of phone solar battery chargers from ebay. Prices start at 15USD, these cheap chargers are probably total garbage, but you can try. Everything is there - rechargeable battery, charging circuits, constant voltage output. Just disassemble it and put its guts into your painting.
You can reduce power requirements by adding some movement sensor to detect people in front of the painting. It may be infrared or microphone (needs less power).
Another idea - you can use photoresistor to measure light and make some adaptive brightness.
And for god sake - remove these wires during exhibition, use any kind of battery!!! :)
Add to your 2 V load at 1800 mWh the power consumption of the RPi. Find a good guess on that at RPi@SE. Given the amount the RPi consumes (20 to 45 Wh a day) your load seems to be negligible for your power supply. So the most simple solution would be to pick a suitable power supply for the RPi and use a switching step down regulator to provide the 2 V.
(Depending on the application the RPi might not be best suited for a portable solution.)
Best Answer
A small 5V latching relay will do it. As it latches, it consumes no power except when changing state.