Reacting sodium metal with aqueous sodium hydroxide, what would happen?
Consider first the solvation of sodium metal in water to give a sodium ion and a hydrated electron.
$$\ce{Na(s) \xrightarrow{\ce{H_2O}} Na^+(aq) + e^-(aq) }$$
The hydrated electron will then act as the reactive species to reduce water to hydrogen atoms and hydroxide anions.
$$\ce{e^-(aq) + H_2O(l) \rightarrow H(aq) + OH^-(aq)}$$
Note that two of these hydrogens come together to form the diatomic hydrogen gas.
$$\ce{2H(aq) \rightarrow H_2(g)}$$
When we multiply the coefficients in the first two equations by two, we are able to consider the net ionic equation through the elimination of the reactive intermediates.
$$\ce{2Na(s) + 2H_2O(l) \rightarrow 2Na^+(aq) + 2OH^-(aq) + H_2(g)}$$
We see that the dissolution of sodium metal in aqueous sodium hydroxide would be affected by Le Chatelier's principle due to the presence of the sodium hydroxide. We would expect there to be a reduction of solubility and rate of dissolution. The greater the initial concentration of sodium hydroxide, the more appreciable the effect.
Given that a solution of sodium hydroxide at $\ce{25 ^{\circ}C}$ is saturated at $\ce{25 M}$, a $\ce{1 M}$ solution would not be expected to have an appreciable effect. Hydrogen gas would remain a product, evidenced by the net equation.
As far as the reaction of molten sodium and molten sodium hydroxide, I expect that is what they were referring to on the Wiki page for sodium oxide when they gave this equation.
$$\ce{2NaOH + 2Na \rightarrow 2Na_2O + H_2}$$
The source given is Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.). Butterworth-Heinemann. ISBN 978-0-08-037941-8.
The solubility of sodium hydroxide was resourced from the Wiki page for sodium hydroxide. The source given appears to be Haynes, CRC Handbook of Chemistry and Physics.
The first set of equations were sourced from Walker, David C. Production of Hydrated Electrons. Canadian Journal of Chemistry. Volume 44. 1966.