As you said, I cannot see the final product of the mechanism because of the logo on the picture, but it looks like the double bond is formed between the N and Carbon C1 from the mechanism diagram. If my guess about the picture is correct, then the picture you've given supports my hypothesis that in the reaction between amphetamine and formaldehyde the imine intermediate has a double bond between N and the methyl carbon and NOT between N and the alpha carbon.
If the mechanism photo you have provided shows something else under the occluding logo, please send a link to where I can see the diagram without the logo, as I would like to see it, however it seems to support what I have said earlier.
However my mechanistic understanding is not top-tier and is certainly lacking in expertise. Do you have any idea on how the above mechanism can generate a double bond between N and the alpha carbon, given that the O that forms the H2O leaving group is attached to the formaldehyde C1 carbon from the mechanism diagram?
As for "reliable data" about the stereospecifity, I have none other than the original Rhodium article purports that the synthesis is stereospecific, as another member said early in the thread on page 1 of replies, here:
https://www.erowid.org/archive/rhodium/chemistry/amphetamine.methylation.html
The above article states that forming imine with d-amphetamine and reducing gives d-methamphetamine. Given that the currently-shown structure (3) is identical to the phenylacetone-methylamine imine, the reduction of which yields a racemic product, it is clear that one of the two must be wrong, no?
Either the amphetamine-formaldehyde imine is different than the phenylacetone-methylamine imine, or the reaction is not stereospecific as is stated in the Rhodium source.
Someone with the adequate resources could perhaps test the reaction with different optical mixtures of amphetamine and report back, however I cannot gather this data.
Thank you for your replies.