$\DeclareMathOperator\GF{GF}$Let $p$ and $p+2$ be twin primes. Let's assume that $-1$ is not a quadratic residue modulo $p$ (and therefore is a Q.R. modulo $p+2$).
Consider the complex numbers $a+bi$ with $a^2+b^2 = 1$ modulo $p$. There are $p+1$ many such numbers and the form a cyclic subgroup of $\GF(p^2)^\times$ with order $p+1$.
Consider the non-zero numbers modulo $p+2$. There are $p+1$ many such numbers and they form an isomorphic cyclic subgroup equal to $\GF(p+2)^\times$ with order $p+1$.
Are there any intrinsic isomorphisms between these two groups?
Notes: I was sort of able to construct an intrinsic map when all the factors of $p+1$ are Fermat primes. See the observation below.
Here are some interesting things I tried that didn't lead anywhere:
Variations of the map $a+bi \to a+bk$ where $k^2 = -1$ modulo $p+2$ looked promising but ultimately didn't work.
The number of solutions to $a^2+b^2 =1$ modulo $p$ and modulo $p+2$ is equal so I looked for maps of form $a+bi \to f(a,b) +g(a,b)k$ where $f$ and $g$ go from modulo $p$ to modulo $p+2$.
Weird observation:
The solutions to $a+bi$, $a^2+b^2=1$ modulo $p$ turn out to be exactly $e^{2\pi i k/p+1}$ modulo $p$ for some $k$ (if $-1$ is a Q.R. mod $p$ then we get solutions of form $e^{2\pi i k/p-1}$). What I mean by this is that if you write down a $p^{\text{th}}$ root of unity exactly as an expression involving radicals and reduce modulo $p$ you will get a solution to $a^2+b^2 =1$ mod $p$.
For example, a third root of unity is $-1/2 + \sqrt{3}i/2$. Now reducing this modulo $p$ will give you an element of norm $1$, $a+bi$ with $a^2+b^2 =1$ mod $p$ and $(a+bi)^3 =1$ (if $3$ divides $p+1$ or $p-1$ in case $-1$ is a Q.R. mod $p$). You can try this out with $p=11$ and $p+2=13$ to see what I mean.
Unfortunately, large roots of unity can't be written down as radicals, so while this counts an intrinsic map, it can't really be computed. It would only be useful for Fermat primes (the roots of unity of Fermat primes can be written as radicals exactly).
