A property of 47 with respect to partitions into five parts Is 47 the largest number which has a unique partition into five parts (15, 10, 10, 6, 6), no two of which are relatively prime?

Yes. Suppose $n>47$.
If $2\mid n$, we can take $(n-8,2,2,2,2),(n-10,4,2,2,2)$, which are distinct partitions for $n\geq 14$.
If $3\mid n$, we can take $(n-12,3,3,3,3),(n-15,6,3,3,3)$, which are distinct partitions for $n\geq 21$.
If $n\equiv 1\pmod 6$, we can take $(n-37,15,10,6,6),(n-43,15,12,10,6)$, which are distinct partitions for $n\geq 55$, and $(21,7,7,7,7),(14,14,7,7,7)$ for $n=49$.
If $n\equiv 5\pmod 6$, we can take $(n-41,15,10,10,6),(n-47,15,12,10,10)$, which are distinct partitions for $n\geq 59$, and $(20,15,6,6,6),(15,12,10,10,6)$ for $n=53$.
(Thanks to Gerhard for pointing out we can finish the argument quickly from what I wrote before)
