= Solution
A sufficient condition is that $T$ be <first-order consistent>, complete, and full in its own <first-order language>. Work in the allowed function-free setting. Since $T$ proves the logically valid <first-order sentence> $\exists x(x=x)$, fullness guarantees at least one constant and a nonempty collection of names.
Form the carrier from <equivalence classes> $[c]$ of <constant symbols> under
$$
c\sim d\quad\Longleftrightarrow\quad T\vdash c=d.
$$
<Logical equality> logic makes this an <equivalence relation>. Interpret a constant by its class and interpret each relation symbol by
$$
R([c_1],\ldots,[c_n])\quad\Longleftrightarrow\quad
T\vdash R(c_1,\ldots,c_n).
$$
Substitution of equals makes this independent of representatives. This defines a <canonical first-order structure> because every carrier element is named by a constant.
The truth lemma, proved by <induction> on <logical formulas> with constants substituted for their <free variables>, says
$$
\mathcal M\models\phi([\vec c])\quad\Longleftrightarrow\quad T\vdash\phi(\vec c).
$$
<Atomic formulas> hold by construction; completeness and <first-order consistency> give the Boolean steps. For the existential step, a witness in the <first-order structure> is a named class, so its truth gives a proved instance and hence a proved <existential sentence>. Conversely a proved <existential sentence> has a constant witness by fullness, and the <induction> hypothesis realizes that instance in the <first-order structure>. The universal case follows by <negation>. Thus $\mathcal M\models T$, proving
$$
\boxed{\text{Consistent complete full theories have canonical first-order structures}.}
$$
The witness condition is needed in the original <first-order language>; adding new constants produces a <canonical first-order structure> in an expanded <first-order language>, which is a different assertion.
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