Suppose $c:=[a_1,a_2,\cdots,a_n]$ is a continued fraction and $T(c):=[a_2,a_3,\cdots,a_n,a_1]$. Is there a simple algebraic relation amongst $c$ and its cyclic transforms $T^i(c),\,\forall i\in\{1,2,\cdots,n-1\}$?
The same question for $c:=[\overline{a_1,a_2,\cdots,a_n}]$.
For the purely periodic c.f. $c=[\overline{a_1,a_2,\cdots,a_n}]$ we have
$c = a_1 + \frac{1}{T(c)}$
so
$T(c) = \frac{1}{c-a_1}$
$T^2(c) = \frac{1}{T(c) - a_2}=\frac{c-a_1}{1+a_1a_2-a_2c}$
$T^3(c) = \frac{1}{T^2(c) - a_3}=\frac{1+a_1a_2-a_2c}{-a_1-a_3-a_1a_2a_3+ (1+a_2a_3)c}$
etc.
Eventually we have $T^n(c) =\frac{P(c)}{Q(c)} = c$ and we can prove by induction that $P(c)$ and $Q(c)$ are linear fractional functions of $c$. This shows that a purely periodic c.f. is a solution to a quadratic equation (and, indeed, this is also true for eventually periodic c.f.s too).