Sort only the odd elements of an array

I want to sort only the odd elements of an array, leaving the even elements unchanged.

I have created the below using LINQ queries, but get an error on this line:

.Select(x => x.value % 2 == 1 ? x.value : dictionary[x.index])

The error is due to there not being an item in the dictionary with key '2'. I.e. the dictionary[x.index] term in the if statement is being evaluated when the criteria for the if statement is false (i.e. when x.value is even).

public class Kata
{
    public static int[] SortArray(int[] array)
    {
        var dictionary = array
                            .Select((value, index) => new { value, index })
                            .Where(x => x.value % 2 == 1)
                            .OrderBy(x => x.value)
                            .ToDictionary(x => x.index, x => x.value);

        var output = array
                            .Select((value, index) => new { value, index })
                            .Select(x => x.value % 2 == 1 ? x.value : dictionary[x.index])
                            .ToArray();

        return new int[1];
    }
}

public class Program
{
    static void Main(string[] args)
    {
        Console.WriteLine(Kata.SortArray(new int[] { 5, 3, 2, 8, 1, 4 }));
    }
}

You can extract odd items, sort them, and finally put them back:

int[] array = new int[] { 5, 3, 2, 8, 1, 4};

// odd items extracted and sorted
using var odd = array
  .Where(item => item % 2 != 0)
  .OrderBy(item => item)
  .GetEnumerator();

for (int i = 0; i < array.Length; ++i) 
  if (array[i] % 2 != 0 && odd.MoveNext()) 
    array[i] = odd.Current; 

Fiddle

Please, note that checking for odd should be item % 2 != 0, not item % 2 == 1, since remainder % returns -1 on negative item.