Every integer $N$ can be represented by a sum of 4 squares $N=a^2+b^2+c^2+d^2$. We usually have more than one representation for a given integer N. For example $7*13=91$ has the following representation written without the exponent 2. $$91=(5,5,5,4),(5,7,4,1),(5,8,1,1),(8,3,3,3),(9,3,1,0)$$
All these representations are equal to the same number but can they be called equivalent if we can transform one into another. We will give just one example. We take the first representation $(5,5,5,4)$ and show that it can be transformed into the second one $(5,7,4,1)$. $$5^2=3^2+4^2=4^2+2^2+2^2+1^2=(4,2,2,1)$$ By substituting $(5,5)$ in the first representation by $(4,2,2,1),(4,2,2,1)$, rearranging we get $$(5,5)=4^2+2^2+2^2+1^2+4^2+2^2+2^2+1^2=4^2+4^2+4^2+1^2+1^2=7^2+1^2=(7,1)$$ if we now substitute $(5,5)$ by its new value $(7,1)$ into the first representation, we get the second one $(5,7,4,1)$. It is clear that we can take the second representation and transform it into the first.
Is this a general result? Is it possible to transform every representation into every other representation for a given integer $N$, and if yes, can it be proven?
If it is possible to always transform one 4-sq representation into another one, then we do not need to find all the representations because it is expensive to calculate all of them. We would in that case only need the first one, then transform it into another one if needed.
I think you're suggesting that a $4$-tuple $[a,b,c,d]$ is equivalent to another $4$-tuple obtained by replacing two of $a,b,c,d$ by some other pair whose sum of squares is the same, the other two remaining unchanged. Presumably the relation is the transitive closure of this, i.e. if $[a,b,c,d]$ is equivalent to $[a',b',c',d']$ and $[a',b',c',d']$ to $[a'',b'',c'',d'']$ then $[a,b,c,d]$ is equivalent to $[a'',b'',c'',d'']$. But, in your example, $(8,3,3,3)$ is not equivalent to any other solution.