I have the following function normalised to 1 on $(0, \infty)$: $$ g(x) = \frac{e^{-\left(\mu + \frac{\sigma^2}{2}\right)} e^{-\frac{(\mu -\text{Log}[x])^2}{2 \sigma ^2}}}{\sigma \sqrt{2 \pi }} $$ which, after multiplying by $\frac{1}{x}e^{\left(\mu + \frac{\sigma^2}{2}\right)}$ gives a log-normal distribution.
The n-th moment of g is $e^{n \mu + \frac{n^2\sigma ^2}{2} + n \sigma^2}$, which again is the same as for the log-normal to the factor $e^{n \sigma^2}$.
It seems like I might be missing some theory here but how can I show that $g(x)$ and the log-normal distribution are essentially the same? I can make them overlap in the plot...
(EDIT) The reason I think $g(x)$ and the log-normal distribution must be somehow related is the following. Suppose I set the first two moments, $m_1$ and $m_2$, of these distributions to same values. For $g(x)$ I obtain the following scale and shape parameters: $$ \mu_g = \log{\frac{m_1^4}{m_2^{3/2}}}\\ \sigma_g = \sqrt{\log{\frac{m_2}{m_1^2}}} $$
For the log-normal distribution I get the following: $$ \mu_{LN} = \log{\frac{m_1^2}{m_2^{1/2}}}\\ \sigma_{LN} = \sigma_g $$
After substituting the above parameters to their respective distributions I calculate the ratio: $$ \frac{g(x)}{f_{LN}(x)} = x \frac{m_2}{m_1^3} \left( \frac{m_1^2}{m_2} \right)^{\log{ \left( \frac{m_2}{m_1^3} x \right)} / \log{ \left( \frac{m_2}{m_1^2} \right)}} = 1 $$ which I find quite surprising especially because the two distribution differ by a factor that is not a constant but depends on $x$. I wonder whether this is a case of a more general rule?