= Solution
We describe a collection of finite multiplication patterns. For finite sets
$$
E\subseteq\{1,\ldots,n\}^3,
\qquad
D\subseteq\{1,\ldots,n\}^2,
$$
write
$$
P_{E,D}(x_1,\ldots,x_n)=
\bigwedge_{(i,j,k)\in E}x_ix_j=x_k
\ \wedge\!
\bigwedge_{(i,j)\in D}x_i\ne x_j.
$$
Call $(E,D)$ $T$-forbidden if the following theory, in $L$ expanded by unary symbols $f_1,\ldots,f_n$, is inconsistent:
* $T$, together with the assertion that every $f_i$ is an automorphism;
* $\forall z\,f_i(f_j(z))=f_k(z)$ for every $(i,j,k)\in E$;
* $\exists z\,f_i(z)\ne f_j(z)$ for every $(i,j)\in D$.
Let $T^*$ contain the group axioms and, for every $T$-forbidden finite pattern, the sentence
$$
\forall x_1\cdots\forall x_n\,\neg P_{E,D}(x_1,\ldots,x_n).
$$
This is a first-order theory in the language of groups.
Suppose $G$ is $T$-good. Choose $M\models T$ and an embedding $\rho:G\hookrightarrow\operatorname{Aut}(M)$. If a tuple $(g_1,\ldots,g_n)$ in $G$ realized a forbidden pattern, interpreting $f_i$ as $\rho(g_i)$ would give a model of its supposedly inconsistent automorphism theory. Thus no such tuple exists, and $G\models T^*$.
Conversely, suppose $G$ is $T$-bad. By part (e), there is a finite set $X=\{g_1,\ldots,g_n\}\subseteq G$ for which $T_G\cap L_X$ is inconsistent. Let
$$
E=\{(i,j,k):g_ig_j=g_k\},
\qquad
D=\{(i,j):g_i\ne g_j\}.
$$
The associated automorphism theory is precisely $T_G\cap L_X$, up to renaming its function symbols, so $(E,D)$ is $T$-forbidden. But $(g_1,\ldots,g_n)$ realizes $P_{E,D}$ in $G$, and therefore $G$ violates the corresponding forbidding axiom. Hence $G\not\models T^*$.
We have proved
$$
\boxed{G\models T^*\quad\Longleftrightarrow\quad G\text{ is }T\text{-good}.}
$$
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