Good news! This going to be cheap! :-)
A simple resistor divider will bring the 12 V down to the 5 V an Arduino can digest. The output voltage can be calculated as
\$ V_{OUT} = \dfrac{R2}{R1+R2} V_{IN}\$
Resistor values in the range of 10 kΩ are a good choice. If your R2 is 10 kΩ then R1 should be 14 kΩ. Now 14 kΩ is not a standard value, but 15 kΩ is. Your input voltage will be 4.8 V instead of 5 V, but the Arduino will see that still as a high level. You also have a bit of headroom in case the 12 V should be a bit too high. Even 18 kΩ will still give you a sufficiently high 4.3 V, but then you have to start thinking about the 12 V a bit too low. Will the voltage still be seen as high? I would stick with the 15 kΩ.
edit
You mention an automotive environment, and then you do need some extra protection. The car's 12 V is never quite 12 V, but most of the time higher, with peaks several volts above the nominal 12 V. (Actually nominal is more like 12.9 V, at 2.15 V per cell.) You can place a 5 V zener diode in parallel with R2, and this should cut off any voltage higher than the zener's 5 V. But a zener voltage varies with the current, and at the low input current the resistors give you it will cut off at lower voltages. A better solution would be to have a Schottky diode between the Arduino's input and the 5 V supply. Then any input voltage higher than about 5.2 V will make the Schottky diode conduct, and the input voltage will be limited to the 5.2 V. You really need a Schottky diode for this, a common P-N diode has a 0.7 V drop instead of the Schottky's 0.2 V, and then the 5.7 V maximum input voltage may be too high.
Better
Michael's optocoupler is a good alternative, though a bit more expensive. You often will use an optocoupler to isolate input from output, but you can also use it to protect an input like you want here.
How it works: the input current light the internal infrared LED, which causes an output current through the phototransistor. The ratio between input and output current is called CTR, for Current Transfer Ratio. The CNY17 has a minimum CTR of 40 %, which means you need 10 mA input for 4 mA output. Let's go for the 10 mA input. Then R1 should be (12 V - 1.5 V) / 10 mA = 1 kΩ. The output resistor will have to cause a 5 V drop at 4 mA, then that should be 5 V / 4 mA = 1250 Ω. It's better to have a bit higher value, the voltage won't drop more than 5 V anyway. A 4.7 kΩ will limit the current to about 1 mA.
Vcc is the Arduino's 5 V supply, Vout goes to the Arduino's input. Note that the input will be inversed: it will be low if the 12 V is present, high when it isn't. If you don't want that, you can swap the position of the optocoupler's output and the pull-up resistor.
edit 2
How doesn't the optocoupler solution solve the overvoltage issue? The resistor divider is ratiometric: the output voltage is a fixed ration of the input. If you have calculated for 5 V out at 12 V in, then 24 V in will give 10 V out. Not OK, hence the protection diode.
In the optocoupler circuit you can see that the right side, which connects to the Arduino's input pin doesn't have any voltage higher than 5 V at all. If the optocoupler is on then the transistor will draw current, I used 4 mA in the example above. A 1.2 kΩ will cause a 4.8 V voltage drop, due to Ohm's Law (current times resistance = voltage). Then the output voltage will be 5 V (Vcc) - 4.8 V across the resistor = 0.2 V, that's a low level. If the current would be lower the voltage drop will be smaller as well, and the output voltage will rise. A 1 mA current, for instance, will cause a 1.2 V drop, and the output will be 5 V - 1.2 V = 3.8 V. The minimum current is zero. Then you don't have a voltage across the resistor, and the output will be 5 V. That's the maximum, there's nothing there which will give you a higher voltage.
What if the input voltage would become too high? You accidentally connect a 24 V battery instead of 12 V. Then the LED current will double, form 10 mA to 20 mA. The 40 % CTR will cause 8 mA output current instead of the calculated 4 mA. 8 mA through the 1.2 kΩ resistor would be a 9.6 V drop. But from a 5 V supply that would be negative, and that's impossible; you can't go lower than 0 V here. So while the optocoupler would very much like to draw 8 mA, the resistor will limit that. The maximum current through it is when the full 5 V is across it. The output will then be really 0 V, and the current 5 V / 1.2 kΩ = 4.2 mA. So whatever power supply you attach the output current won't go higher than that, and the voltage will stay between 0 V and 5 V. No further protection needed.
If you expect overvoltage you'll have to check if the optocoupler's LED can handle the increased current, but the 20 mA will not be a problem for most optocouplers (they're often rated at 50 mA maximum), and besides, that's for double input voltage, which probably won't happen IRL.
The key difference between relays and MOSFETs in this context is that MOSFETs do not have a linear transfer function for signals passing through them.
In the case of the relay shown in the question, the signal from the active video source passes through the relay contacts essentially unmolested, pretty much regardless of signal polarity or content, like through any other electrical switch contacts.
With MOSFETs, the signal would need to be biased to stay precisely within the linear portion of the MOSFET's conduction curve when the MOSFET is in conducting state. A DC bias voltage must be added to the incoming signal (assuming analog video signal), for this purpose.
The transfer characteristic of a MOSFET for small signal operation looks as below. See this paper for further insight.
As seen here, the MOSFET's conduction reduces sharply as input voltage drops towards zero. What is not shown in the graph is that once the input signal drops to a sufficiently negative value, the MOSFET's body diode begins to conduct, regardless of the voltage at the MOSFET gate.
So the entire signal needs to be biased and constrained to stay within the linear portion around the point indicated as Q
. This might not go down so well at the receiving end, if the receiver is not designed to work with a DC biased video signal.
In addition, a MOSFET's conduction behavior changes with frequency of signal, much more so than a metal-metal contact such as a MOSFET. Hence, depending on the specific MOSFET being considered, and the frequency of the signal, this too would be a concern.
In other words, a MOSFET is not a suitable replacement for a relay in this application.
Even solid state relays, unless specifically designed for video signal transmission, will not work, because they too contain semiconductor switching elements, which will have similar non-linear behavior as indicated above for MOSFETs. So, not an option.
Best Answer
To understand the rationale of AC/DC and DC-only configurations, MOSFET body diode should be considered. In power MOSFETs bulk is connected to source, and that creates a diode in parallel with the MOSFET channel.
The picture below is from the datasheet of IRF510. It's a generic power N-channel MOSFET.
The picture below is from the datasheet of G3VM (another SSR). It also shows body diodes. [Highlighted by me.]
Due to body diode, the N-ch MOSFET can "open the switch" only when body diode is reverse biased. Drain has to be at a higher potential than the source. Otherwise, the diode will conduct, and MOSFET becomes a switch, which doesn't open.
But what to do with the AC? During one part of the AC waveform, the body diode will be forward biased, during another part of the waveform, it will be reverse-biased. This is why the second MOSFET is used. The MOSFETs are connected in series such that their diodes are anti-parallel. They block each-other, and the MOSFETs can "open the switch" at all times. This capability to handle the AC comes at a cost, though. The two MOSFETs are in series and the losses are double.
The LCA717 datasheet doesn't show body diodes. But the body diodes are there nonetheless.
The designers of LCA717 decided to connect the MOSFET sources to the pin. It makes this SSR more versatile. It allows to connect the MOSFETS both: in series (AC/DC configuration), or in parallel (DC-only configuration). Notice that in DC-only configuration, the body diodes are in parallel too, and both of them can become forward biased. Notice that DC mode can handle twice the current.
On the other hand, the designers of the G3VM decided not to connect the MOSFET sources to the pin. Perhaps, they have done that to have fewer pins, in order to keep the device more compact and to keep the cost lower.