I claim that it is always possible to make all of the integers equal to each other. To see this, consider the allowable transformations

$(a,b,c,d,e,f)\mapsto(a+b,a+b,c,d,e,f)$,

$(a_1,\dots,a_6)\mapsto(a_{\sigma(1)},\dots,a_{\sigma(6)})$, and

$(a,b,c,d,e,f)\mapsto(a/n,b/n,c/n,d/n,e/n,f/n)$ whenever $a/n,b/n,c/n,d/n,e/n,f/n$ are all integers.

We shall write $(a,b,c,d,e,f)\mapsto^*(g,h,i,j,k,l)$ if $(g,h,i,j,k,l)$ can be obtained from $(a,b,c,d,e,f)$ by using a sequence of allowable transformations. We observe that if $(a,b,c,d,e,f)\mapsto^*(1,1,1,1,1,1)$, then can transform $(a,b,c,d,e,f)$ into $(n,n,n,n,n,n)$ for some $n$ using only transformations $1$ and $2$, but transformation 3 makes it easier to use induction to show that we can always get $(a,b,c,d,e,f)\mapsto(1,1,1,1,1,1)$.

We observe that $(a,b,c,d,e,f)\mapsto^*(a+b+c+d,a+b+c+d,a+b+c+d,a+b+c+d,e+f,e+f)$.
Now set, $\alpha=a+b+c+d,\beta=e+f$. I claim that $(\alpha,\alpha,\alpha,\alpha,\beta,\beta)=(0,0,0,0,0,0)$ or $(\alpha,\alpha,\alpha,\alpha,\beta,\beta)\mapsto^*(1,1,1,1,1,1)$ by induction on the square of the $\ell^2$ norm $4\alpha^2+2\beta^2$ of $(\alpha,\alpha,\alpha,\alpha,\beta,\beta)$.

Case 0: $\alpha=0,\beta=0$. This is trivial.

Case 1: $\alpha=0,\beta\neq 0$. In this case,
$(\alpha,\alpha,\alpha,\alpha,\beta,\beta)\mapsto^*(\beta,\beta,\beta,\beta,\beta,\beta)\mapsto^*(1,1,1,1,1,1)$.

Case 2: $\alpha\neq 0,\beta=0.$

$(\alpha,\alpha,\alpha,\alpha,\beta,\beta)\mapsto^*(\alpha,\alpha,\alpha,\alpha,\alpha,\alpha)\mapsto^*(1,1,1,1,1,1)$.

Case 3: $\alpha\neq 0,\beta\neq 0$, $\alpha$ is even.

If $\alpha$ is even, then
$(\alpha,\alpha,\alpha,\alpha,\beta,\beta)\mapsto^*(\alpha,\alpha,\alpha,\alpha,2\beta,2\beta)\mapsto^*(\alpha/2,\alpha/2,\alpha/2,\alpha/2,\beta,\beta)$ and
$(\alpha/2,\alpha/2,\alpha/2,\alpha/2,\beta,\beta)\mapsto^*(1,1,1,1,1,1)$ by the induction hypothesis.

Case 4: $\alpha\neq 0,\beta\neq 0$, $\beta$ is even.
$(\alpha,\alpha,\alpha,\alpha,\beta,\beta)\mapsto^*(2\alpha,2\alpha,2\alpha,2\alpha,\beta,\beta)\mapsto^*(\alpha,\alpha,\alpha,\alpha,\beta/2,\beta/2)$, and
$(\alpha,\alpha,\alpha,\alpha,\beta/2,\beta/2)\mapsto^*(1,1,1,1,1,1)$ by the induction hypothesis.

Case 5: $\alpha\neq 0,\beta\neq 0$, $\alpha,\beta$ are both odd.

Let $\alpha=2\gamma+1,\beta=2\delta+1.$ Then $(\alpha,\alpha,\alpha,\alpha,\beta,\beta)
\mapsto^*(2(2\gamma+1),2(2\gamma+1),2(\gamma+\delta+1),2(\gamma+\delta+1),2(\gamma+\delta+1),2(\gamma+\delta+1))$
$\mapsto^*(2\gamma+1,2\gamma+1,\gamma+\delta+1,\gamma+\delta+1,\gamma+\delta+1,\gamma+\delta+1)=(\alpha,\alpha,\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2}).$

If $\alpha=\beta$, then clearly $(\alpha,\alpha,\alpha,\alpha,\beta,\beta)
\mapsto^*(1,1,1,1,1,1)$, and if $\alpha\neq\beta$, then we may use the induction hypothesis to conclude that $(\alpha,\alpha,\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2})\mapsto^*(1,1,1,1,1,1).$

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