The title says it all.
Let $\sigma = \sigma_{1}$ denote the classical sum-of-divisors function. That is, $$\sigma(N) = \sum_{d \mid N}{d}.$$ For example, $\sigma(10) = 1 + 2 + 5 + 10 = 18$.
Here is my question:
Is there an interpretation for why $\sigma(x) = x \iff x = 1$, involving notions that can be extended outside elementary number theory?
Things I Have In Mind
(1) If $\sigma(x) = x$, then $$\gcd(x, \sigma(x)) = x = \sigma(x).$$ How then do I show that $x=1$ (using only the concept of the greatest common divisor, (possibly) the definition of the $\sigma$ function, and no other)?
(2) If $\sigma(x) = x$, then $x$ is said to be a fixed point of $\sigma$. Using concepts from functional analysis (and possibly, the definition of the $\sigma$ function), how do we then show that $x=1$?
(3) If $\sigma(x) = x$, then since $\sigma$ is a function, we can apply $\sigma$ to both sides repeatedly to obtain $$x = \sigma(x) = \sigma(\sigma(x)) = \sigma(\sigma(\sigma(x))) = \sigma(\sigma(\sigma(\sigma(x)))) = \sigma(\sigma(\sigma(\sigma(\sigma(x))))) = \ldots.$$ This suggests a possible connection to recursion theory. In particular, how do we then show that $x=1$, using only such a concept?
$$\sigma(n) = \sum_{d\mid n} d \geq \sum_{d\mid n} 1 = d(n) \tag{1}$$ hence $\sigma(n)=1$ implies $d(n)=1$, but every natural number $n\geq 2$ has at least two distinct divisors, $1$ and $n$. Assuming $s>2$, from $$ \sum_{n\geq 1}\frac{\sigma(n)}{n^s} = \zeta(s)\,\zeta(s-1) \tag{2}$$ and summation by parts it follows that $$ \frac{1}{N}\sum_{n=1}^{N}\sigma(n) = \frac{\pi^2 N}{6} + O(\sqrt{N})\tag{3} $$ hence $\sigma(n)=1$ may hold only for $n\in[1,N_0]$ for a fixed $N_0$, since $(3)$ gives $\sigma(N)=\frac{\pi^2}{3}N+O(\sqrt{N})$, on average.