june

There are two correct answers here:

Under C99, C11, and C17 (GCC/Clang flags "-std=c99", "-std=c11", or "-std=c17"), the answer is "this won't compile for some other reason".

Under C89, C95, and C23 (GCC/Clang flags "-std=c89", "-std=iso9899:199409", and "-std=c23"), the answer is int.

Why? A few reasons:

auto is, in fact, a keyword. It has been since C89, and since K&R before that, and since B before that. It is a storage-class specifier (like static or thread_local or constexpr) and means "automatic storage duration"; AKA "this variable is freed at the end of the enclosing scope", AKA "this is a stack variable". Automatic storage is the default storage class for local variables (of course), and also only valid for local variables, so prior to C23 specifying it was always redundant.

Prior to C99, variable declarations without a type were implicitly given the type int.¹ Thus, in C89 and C95, this declaration is equivalent to auto int x = 'a';

From C99 until C23, this won't compile, but only because variable declarations without a type are illegal. auto is still a legal storage specifier. The relevant GCC/Clang diagnostic is -Wimplicit-int.

In C23 this declaration becomes legal once again, because a variable declared with the auto storage specifier and without a type (a so-called "direct declarator") now performs type inference. This is equivalent to writing auto typeof('a') x = 'a';

"But wait," you might ask, "why is x an int, if its type is inferred from its initializer? Isn't 'a' a char?" Nope! That's the case in C++, but confusingly, character literals in C have the type int, and thus so does x. We would have to write auto x = (char)'a'; to declare x to be a char.