The comparison is utterly meaningless.
Consider, for instance, fitting iid data $\mathbb{x} = (x_1, \ldots, x_n)$ to a location-scale family having a continuous pdf
$$f(x; \mu, \sigma) = f\left(\frac{x-\mu}{\sigma}\right) / \sigma.$$
The log likelihood for these data equals
$$\Lambda(\mathbb{x}; \mu, \sigma) = \sum_{i=1}^n \log \left(f\left(\frac{x_i-\mu}{\sigma}\right)/\sigma\right) = -n \log(\sigma) + \sum_{i=1}^n \log f\left(\frac{x_i-\mu}{\sigma}\right).$$
Because $\frac{x-\mu}{\sigma} = \frac{\alpha x-\alpha\mu}{\alpha\sigma}$ for any positive real number $\alpha$, $(\hat{\mu}, \hat{\sigma})$ maximize the log likelihood for $\mathbb{x}$ if and only if $(\alpha \hat{\mu}, \alpha \hat{\sigma})$ maximize the log likelihood for a corresponding dataset $(\alpha x_1, \ldots, \alpha x_n)$ obtained by rescaling the original data. This is tantamount to a change in their units of measurement, so the "fit" to the model cannot be any better or worse. However,
$$\Lambda(\mathbb{x}; \hat{\mu}, \hat{\sigma}) = n\log(\alpha) + \Lambda(\alpha\mathbb{x}; \alpha\hat{\mu}, \alpha\hat{\sigma}).$$
It follows that we may change the optimal value of the log likelihood by any arbitrary amount $n\log(\alpha)$ by means of an appropriate choice of $\alpha$ without changing the fit (at least not in any statistically reasonable sense). For instance, if your data are lengths measured in meters and you were to re-express those lengths in kilometers, you would thereby increase the maximum log likelihood by $n\log(1000)$ without changing the "fit."
Because both AIC and BIC differ from the optimal value of $\Lambda$ by functions depending only on $n$, which does not change, the same conclusion holds for them, too.
