Replace occurrences of two characters with each other in a string

Use an NSScanner to move through the string and substitute each character as it is found. This way, all substitutions are done in one pass and you're never looking at a position twice.

NSMutableString * fixedUpString = [NSMutableString string];
NSScanner * scanner = [NSScanner scannerWithString:origString];
NSCharacterSet * subCharacters = [NSCharacterSet characterSetWithCharactersInString:@"*/"];

while( ![scanner isAtEnd] ){

    // Pick up other characters.
    NSString * collector;
    if( [scanner scanUpToCharactersInSet:subCharacters intoString:&collector] ){
        [fixedUpString appendString:collector];
    }

    // This can easily be generalized with a loop over a mapping from
    // found characters to substitutions
    // Check which one we found
    if( [scanner scanString:@"*" intoString:nil] ){

        // Append the appropriate substitution.
        [fixedUpString appendString:@"/"];
    }
    else /* if( [scanner scanString:@"/" intoString:nil] ) */  {
        [fixedUpString appendString:@"*"];
    }
}

fixedUpString now contains the substituted content.

As I noted in the comment, this can be generalized very easily to any number of substitutions:

NSDictionary * substitutions = @{ @"a" : @"z", @"b" : @"y", ... };
NSCharacterSet * keyChars = [NSCharacterSet characterSetWithCharactersInString:[[substitutions allKeys] componentsJoinedByString:@""]];

...

// Check which one we found
for( NSString * keyChar in [substitutions allKeys] ){

    if( [scanner scanString:keyChar intoString:nil ){
        [fixedUpString appendString:substitutions[keyChar]];
        break;
    }
}

I had to have a go at this without using the intermediary & character, and though this is certainly more convoluted, this also appears to work:

int main(int argc, char *argv[])
{

    NSString *s = @"1*2/3*4*5*6*7*8/2";

    NSArray *stars = [s componentsSeparatedByString:@"*"];
    NSMutableArray *slashes = [NSMutableArray array];
    for (NSString *star in stars)
    {
        [slashes addObject:[star componentsSeparatedByString:@"/"]];
    }

    NSMutableArray *newStars = [NSMutableArray array];
    for (NSArray *slash in slashes)
    {
        [newStars addObject:[slash componentsJoinedByString:@"*"]];
    }

    NSString *newString = [newStars componentsJoinedByString:@"/"];

    NSLog(@"%@", newString);

    return 0;
}

Output:

1/2*3/4/5/6/7/8*2

Here's my take on a solution, since I could not find a builtin method that would do what you are looking for.

NSString *string = @"a*b/c";
NSDictionary *swapings = @{@"*" : @"/", @"/" : @"*"};
NSString *newString = [self swap:swapings inString:string];
NSLog(@"'%@' became '%@'", string, newString);

# ... somewhere in self :

-(NSString *)swap:(NSDictionary *)swapings inString:(NSString *)string {
    NSMutableArray *letters = [NSMutableArray array];
    for (int i = 0; i < [string length]; i++) {
        NSString *letter = [string substringWithRange:NSMakeRange(i, 1)];
        [letters addObject:letter];
    }

    for (int i=0; i<letters.count; i++) {
        NSString *letter = [letters objectAtIndex:i];
        for(NSString *token in [swapings allKeys]) {
            if ([letter isEqualToString:token]) {
                letter = [swapings valueForKey:token];
                break;
            }
        }
        [letters replaceObjectAtIndex:i withObject:letter];
    }

    return [letters componentsJoinedByString:@""];
}