= Solution
Let $q_0$ be the <conductor of a Dirichlet character> and $\chi_0$ the inducing primitive even character. Removing the Euler factors absent from $L(s,\chi)$ gives the <imprimitive Dirichlet L-function Euler correction>
$$
L(s,\chi)=P(s)L(s,\chi_0),\qquad
P(s)=\prod_{\substack{p\mid q\\p\nmid q_0}}(1-\chi_0(p)p^{-s}).
$$
The primitive <functional equation> therefore gives
$$
L(s,\chi)=\varepsilon_{\chi_0}\left(\frac{q_0}{\pi}\right)^{1/2-s}
\frac{\Gamma((1-s)/2)}{\Gamma(s/2)}P(s)L(1-s,\overline\chi_0).
$$
Equivalently, replace the final primitive function by $L(1-s,\overline\chi)/P_{\overline\chi}(1-s)$, interpreted as a <meromorphic> identity with removable values handled by continuation. It is the <conductor of a Dirichlet character> $q_0$, rather than the possibly inflated modulus $q$, that enters the gamma factor and root number. The <complex conjugation> bar in the original PDF is lost in the converted TeX.
The zeros are those of $L(s,\chi_0)$ together with the zeros of the finite <Euler product>, and multiplicities add. Since $|\chi_0(p)|=1$ at each extra <prime>, an extra factor vanishes at the imaginary points determined by
$$
p^{-s}=\overline{\chi_0(p)}.
$$
Each extra <prime> creates infinitely many such points. Its nonzero-imaginary points are not zeros of the primitive function: the <functional equation> and <nonvanishing of Dirichlet L-functions on the line one> exclude them. Thus the zero sets are identical precisely when every <prime> dividing $q$ already divides $q_0$, making $P=1$. Increasing prime-power exponents alone can make a character imprimitive without changing its L-function. If $q_0=1$, the primitive function is zeta; the same Euler correction applies, with its <pole> at one retained.
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