A comment over on tex.SE made me wonder. The statement is essentially:
If I can write a compiler for language X in language X, then X is Turing-complete.
In computability and formal languages terms, this is:
If $M$ decides $L \subseteq L_{\mathrm{TM}}$ and $\langle M \rangle \in L$, then $F_L = \mathrm{RE}$.
Here $L_{\mathrm{TM}}$ denotes the language of all Turing machine encodings and $F_L$ denotes the set of functions computed by machines in $L$.
Is this true?
The informal statement is not true, as shown by the following programming language. Any string of, say, ASCII characters is a valid program and the meaning of every program is, "Output a program that just outputs a copy of its input." Thus, every program in this language is a compiler for the language but the language is not Turing-complete.
I'm not sure if your "computability theory version" is equivalent but it is also not true. By Kleene's second recursion theorem, for any coding of Turing machines, there is a TM that accepts its own coding and rejects all others.1 This machine is a counterexample to the proposition. More concretely, we can achieve the result by choosing a coding. For example, let every odd number code the machine $M$ defined by "If my input is odd, accept it; otherwise, reject" and let the number $2x$ code the machine coded by $x$ in your own favourite coding scheme for Turing machines. $\langle M\rangle$ is in the language $L$ accepted by $M$ but $F_L$ is not Turing complete.
1 Kleene's second recursion theorem says that, for any enumeration $(\phi_i)_{i\geq 0}$ of the partial recursive functions (i.e., for any coding of programs as integers), and any partial recursive function $Q(x,y)$, there is an integer $p$ such that $\phi_p$ is the function that maps $y$ to $Q(p,y)$. So, in particular, let $Q$ be the function that accepts if $x=y$ and rejects otherwise. By the theorem, there is an integer $p$ that codes the program $\phi_p(y) = Q(p,y)$. That is, $\phi_p$ accepts its own coding $p$ and rejects all other inputs.
