Batteries connected in series provide the sum of their voltage but at the current of the lowest rated battery.
Batteries in parallel (of the same voltage) provide the same voltage as one battery, but at the current capacity of all battery current capacities summed.
Your amplifier is rated at 20V/3A which means it draws 3A most likely at maximum volume. I would expect it draws much less at lower volumes. However, I will provide calculations based on the highest current draw.
"D" cells (batteries) store more energy (18000+ mAh) than "AA" cells (~2750 mAh), and therefore will last longer than "AA" cells for a given load.
If you connected 14 "D" cells in series, that would provide 21 volts (a 5% voltage increase should be tolerated by your amplifier, and as the batteries drain, they will be significantly lower than this voltage).
To address the current draw, you will need to connect additional sets of 14 "D" cells in parallel. Each set connected will increase the current capacity. With only one set, you have a theoretical maximum of 18 amp-hours, which means you could operate a 3A device for 6 hours. However, if you look at the datasheet for "D" cells, the mAh capacity is significantly reduced at high current draw. For example, at just 0.5 A, the mAh capacity is given as ~1500 mAh. This means that your 3A amplifier would only run for a half hour! It gets worse - because you are drawing much more than 0.5A, the life of the batteries is much less. In fact I would not expect the unit to work properly with only one set of batteries in series.
Ideally you would connect enough sets in parallel so that you are only drawing 25mA from each set, which gives the best-case mAh capacity. But this would require that you connect an impractical 120 sets; a total of 1680 "D" cells!
With 6 groups of 14 cells, each set would provide 500mA, and come much closer to providing the 1500 mAh capacity shown on the datasheet. I would expect a run time of about a half hour then.
It is highly likely that you can use another 5V power supply.** - either for individual cameras or for a group of cameras.
- Using a 2 wire power pair only - while it would be possible to customise a device/supply pair so that they would only work together* this would be extremely unusual, and D -Link are not known for making it hard to use their equipment in this manner (unlike some other suppliers).
[* eg the device could signal the supply by modulating the current drawn - and/or the supply could wake up the device by voltage signalling. Doable but extremely unlikely. As two examples only: Apple go out of their way to make it hard for other people to interface with their products, and some Dell systems do. ]
What is the camera model?
What is the camera current draw mean/peak?
If operating over a significant distance, voltage drop can be an issue, especially at low voltages and higher currents. Several means of addressing this exist.
As Boa suggested, use adequately thick wires.
"How thick" is needed depends on current draw and circuit length.
Using a supply voltage at the upper tolerable limit for the device will allow close devices to "survive" the voltage and allow greater distance to the end device before voltage is too low.
Adding a capacitor at the feed point for each camera will almost certainly allow lower voltage operation that if not used. A 100 uF electrolytic at say 10V would help and larger or much larger would not be amiss.
A good quality 100 uF 10V aluminium electrolytic capacitor can be bought for $US0.13 each in 10 quantity
and a [1000 uF 10V capacitor(https://www.digikey.com/product-detail/en/panasonic-electronic-components/ECA-1AM102/P5127-ND/244986) for $US0.30 in 10 quantity
Adding a regulator at each camera or close proximity group of cameras and using a somewhat higher voltage power supply allows for voltage drop in the cable.
Regulators are not complex but do need a small degree of 'design'.
Some regulators have more demanding input and output capacitor requirements than others.
Unless specifically intended as "low drop out regulators" a regulator will often drop typically 1.5 to 3 volts across the regulator in normal operation. In this application you may thus need 5V supply + say 2V wire drop + 2V say regulator "dropout voltage" = 9V or more feed voltage. An LDO (low dropout" regulator will have well under 1 V dropout voltage in normal operation - under 0.1V in some devices.
As an example only - the LD1117 has aa dropout voltage of up to 1 V, max current of 800 mA, and costs around $US0.75 in 10 quantity.
Fire risk:
Note that at the currents and voltages concerned fire is not a major risk if sensible construction methods are used - but also not a nonexistent one. At say 10V feed x 5A supply you have 50 Watts available. You can definitely set things alight at that power level if you are clever enough to short the supply in a close-to-supply portion of the circuit. Common sense should avoid this.
Best Answer
Yes, you are correct!
You are making a parallel circuit, so:
$$V_\text{total} = V_1 = V_2 = V_n$$
$$I_\text{total}=I_1+I_2+I_n$$
The current you have to sum, so if one camera needs 2A, the total current will be:
Number of Cameras times 2A.
The voltage is the same for every camera, so you need ensure 7.4V.
You need to do something like this:
Where V its you power supply and each R are your cameras.