I found the solution and added it as an appendix to a paper that has just been published.
I add the explanation here too.
Let us recap the problem. We have a continuous-time signal $ x\left(t\right) $ that is sampled with period $ T_1 $, getting the discrete-time sequence $ x(nT_1) $. A continuous-time ZOH signal $ x_{T_1}^0(t) $ with period $ T_1 $ is then produced from the sequence $ x(nT_1) $. The ZOH signal is then resampled with a period $ T_2 $ such that $ T_1/T_2 = N $ (integer), getting the sequence $ x_{T_1}^0(nT_2) $.
Let $ X(f) $ be the FT of $ x\left(t\right) $, and $ X_1(f) $ the DTFT of the sequence $ x(nT_1) $. It is a well known result that:
$$
X_1(f) = \frac{1}{T_1}\sum_n X\left(f-nF_1\right),
$$
where we assume that $ X(f) $ is band-limited and the sampling rate $ F_1 = 1/T_1 $ fullfills the Nyquist condition. Let us call $ X_1^0\left(f\right) $ the FT of $ x_{T_1}^0(t) $. It is also well known that:
$$
\tag{1}\label{eq_zoh}
X_1^0\left(f\right) = T_1e^{-i\pi f T_1} \mathrm{sinc}\left(fT_1\right) X_1(f) = e^{-i\pi f T_1} \mathrm{sinc}\left(fT_1\right) X(f),
$$
where the second equivalence is true only for $ \left|f\right| < F_1/2 $.
If $ X_1^0\left(f\right) $ is known, $ X(f) $ can be easily reconstructed by inverting (\ref{eq_zoh}). But in general $ X_1^0\left(f\right) $ is not immediately known from the data. Our problem is thus how $ X(f) $ could be reconstructed, given the sequence $ x_{T_1}^0(nT_2) $ and its respective DTFT $ X_2^0\left(f\right) $. The sequence $ x(nT_1) $ (and thus $ X_1\left(f\right) $) could be simply obtained by decimating $ x_{T_1}^0(nT_2) $, but we want to keep all the samples, since, in general, when some noise is present, this yields better results due to averaging over the noise.
Let us define the zero-padded sequence:
$$
x_{zp}\left( {m{T_{2}}} \right) = \left\{ {\begin{array}{*{20}{c}}
{{x}\left( {n{T_1}} \right)}&{m = nN}\\
0&{{\rm{otherwise}}}
\end{array}} \right.,
$$
and let us call $ X_{zp}\left(f\right) $ its DTFT. It is well known that $ X_{zp}\left(f\right) = X_{1}\left(f\right) $ for every value of $ f $. The sequence $ x_{T_1}^0(nT_2) $ can be rewritten as a sum of translated zero-padded sequences:
$$
x_{T_1}^0(nT_2) = \sum_{k=0}^{N-1} x_{zp}\left( {\left(m-k\right){T_{2}}} \right).
$$
Thus, $ X_2^0\left(f\right) $ can be expressed as:
$$
X_2^0\left(f\right) = X_1\left(f\right) \sum_{k=0}^{N-1} e^{-i 2 \pi f k T_2} = X_1\left(f\right) e^{-i\pi f\left(N-1\right)T_2} \frac{\sin\left(N\pi f T_2\right)}{\sin\left(\pi f T_2\right)}.
$$
But:
$$
e^{-i\pi f\left(N-1\right)T_2} \frac{\sin\left(N\pi f T_2\right)}{\sin\left(\pi f T_2\right)} = e^{-i\pi f\left(T_1-T_2\right)} \frac{T_1}{T_2} \frac{\mathrm{sinc}\left(fT_1\right)}{\mathrm{sinc}\left(fT_2\right)}.
$$
Hence, given $ X_2^0\left(f\right) $, $ X_1\left(f\right) $ can be reconstructed as:
$$
X_1\left(f\right) = X_2^0\left(f\right) e^{i\pi f\left(T_1-T_2\right)} \frac{T_2}{T_1} \frac{\mathrm{sinc}\left(fT_2\right)}{\mathrm{sinc}\left(fT_1\right)}.
$$
The last result, formally derived, is the same that I had already figured out empirically and intuitively.
