The following is a string expression which I think does what you describe:
StringReplace[data,
"N" ~~ d : DigitCharacter ~~ x : Except[DigitCharacter] :>
"N0" <> d <> x]
Note that StringReplace does automatically map onto lists of strings (which is documented), so you don't need to Map it onto the list of strings data.
If you prefer regular expressions here is the same expressed with regular expressions:
StringReplace[data, RegularExpression["N(\\d)([^\\d])"] :> "N0$1$2"]
These do both work for your example data, but won't replace when the pattern is at the end of a string, like in "N1 N2 G01 X35 Y51 N3", if you want to also handle those, that can be either achieved with an additional rule:
StringReplace[data, {
"N" ~~ d : DigitCharacter ~~ x : Except[DigitCharacter] :>
"N0" <> d <> x,
"N" ~~ d : DigitCharacter ~~ EndOfString :> "N0" <> d
}]
StringReplace[data, {
RegularExpression["N(\\d)([^\\d])"] :> "N0$1$2",
RegularExpression["N(\\d)$"] :> "N0$1"
}]
or with a slightly more complicated pattern:
StringReplace[data,
"N" ~~ d:DigitCharacter ~~ x:(Except[DigitCharacter] | EndOfString) :>
"N0" <> d <> x]
StringReplace[data,RegularExpression["N(\\d)([^\\d]|$)"] :> "N0$1$2"]
StringReplace[#, "N" -> "N0"] & /@ data? – corey979 yesterday(StringReplace[#, "N" ~~ d : DigitCharacter ~~ WordBoundary :> "N0" <> d]) & /@ data– alancalvitti yesterday