= Solution
Take $T_f$ to consist of $T$ together with the following sentences.
* $f$ is injective and surjective:
$$
\forall x\,\forall y\,(f(x)=f(y)\Rightarrow x=y),
\qquad
\forall y\,\exists x\,f(x)=y.
$$
* For every $n$-ary operation symbol $\omega$ of $L$,
$$
\forall x_1\cdots\forall x_n\,
f(\omega(x_1,\ldots,x_n))
=\omega(f(x_1),\ldots,f(x_n)).
$$
This includes $f(c)=c$ for each constant symbol $c$.
* For every $n$-ary relation symbol $R$ of $L$,
$$
\forall x_1\cdots\forall x_n\,
\bigl(R(x_1,\ldots,x_n)\Longleftrightarrow
R(f(x_1),\ldots,f(x_n))\bigr).
$$
The first pair of axioms makes the interpretation $\vartheta$ a bijection. The remaining schemes say exactly that it preserves every operation and relation, so $(M,\vartheta)\models T_f$ exactly when $M\models T$ and $\vartheta\in\operatorname{Aut}(M)$.
Finally, $T$ is consistent and hence has a model $M$ by part (b). Expanding $M$ by interpreting $f$ as the identity automorphism gives a model of $T_f$, so $T_f$ is consistent.
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