The next colour

JavaScript, 68 bytes

s=>'RedOrangeYellowGreenBlueIndigoVioletRed'.match(s+'(.[a-z]*)')[1]

For input "Red", this function first construct an RegExp /Red(.[a-z]*)/ to match the string 'RedOrangeYellowGreenBlueIndigoVioletRed' and then return the first capture result.

f=
s=>'RedOrangeYellowGreenBlueIndigoVioletRed'.match(s+'(.[a-z]*)')[1]

document.write('<table><tr><th>Input<th>Output')
for(i='Red';;){
document.write(`<tr><td>${i}<td>${i=f(i)}`);
if(i=='Red')break;
}

Perl 5 -p, 58 57 bytes

#!/usr/bin/perl -p
$_={(Red,Orange,Yellow,Green,Blue,Indigo,Violet)x2}->{$_}

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Now that the challenge has been changed to be cyclic the regex solution

say RedOrangeYellowGreenBlueIndigoVioletRed=~/$_(.[a-z]+)/

isn't optimal anymore (due to the double Red)

Also 57 bytes:

#!/usr/bin/perl -p
$_=(Indigo,Blue,Violet,Yellow,Orange,Red,Green)[ord>>2&7]

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Python, 79 bytes

z="Red Orange Yellow Green Blue Indigo Violet".split()*2
dict(zip(z,z[1:])).get

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Handles Violet -> Red. The desired function is given anonymously in the second line.

80 bytes

lambda x:"Red Orange Yellow Green Blue Indigo Violet Red".split(x)[1].split()[0]

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