`{'x', 'y', 'z', '\0'}` does not have a type by itself at all, but it's valid syntax to use it to initialize various structs and arrays - some of those will have the size you are looking for, depending on which type of array or struct you choose to initialize with that.
These following two are incidentally valid whole expressions and have well defined types of char[4] and int[4] respectively. They have different sizes, as is expected from their types.
(char[]){'x', 'y', 'z', '\0'}
(int[]){'x', 'y', 'z', '\0'}
They are literally defined as "characters". sizeof(char) is always 1.
Your confusion (besides the pointer thing) is that 'x' is a funny way to write an int, not a char.
int numbers[] = {1, 2, 3}; // sizeof(numbers) = 12
Yes, this might be it. So the way to get a `char` value that contains "c" is to use type coercion and write it as `(char) 'c'`. This changes the representation in memory so that it now takes up only one byte rather than four, right?
Its size is one byte -- but the size of an expression isn't really relevant, since it's (conceptually) not stored in memory.
You can assign the value of an expression to an object, and that object's size depends on its declared type, not on the value assigned to it. The cast is very probably not necessary.
char c1 = 'c'; // The object is one byte; 'c' is converted from int to char
int c2 = 'c'; // The object is typically 4 bytes (sizeof (int))
But apart from that, I would expect `{'x', 'y', 'z', '\0'}` to have size 16 rather than size 4 because it consists of four character literals which each have size 4 on my machine.