Sequences or 'chains' of adjoint functors Suppose we have (some categories and some functors such that) $F_1$ is left adjoint to $G_1$, $G_1$ left adjoint to $F_2$, $F_2$ left adjoint to $G_2$.
Will $F_1$ then be equal to $F_2$ (and $G_1$ to $G_2$)? 
Or can there exist arbitrarily long 'chains' of adjoint functors that do not trivialize? 
 A: Yes, such infinite chains indeed exist. 
Consider an additive category ${\mathscr A}$, its category of ${\mathbb Z}$-graded chain complexes $\text{Ch}({\mathscr A})$ and the associated forgetful functor ${\mathcal V}: \text{Ch}({\mathscr A})\to{\mathscr A}^{\mathbb Z}$. A left adjoint ${\mathcal L}: {\mathscr A}^{\mathbb Z}\to\text{Ch}({\mathscr A})$ to ${\mathcal V}$ is given by sending some $X_{\ast}\in\text{Obj}({\mathscr A}^{\mathbb Z})$ to the chain complex ${\mathcal L}(X) := \Sigma^{-1}\text{Cone}((X_{\ast},0)\xrightarrow{\text{id}} (X_{\ast},0))$. In words, we first equip $X_{\ast}$ with the trivial differential and then form the cone over it. Explicitly,
$${\mathcal L}(X)\quad\ \equiv\quad ...\to X_{n-1}\oplus X_{n}\xrightarrow{\begin{pmatrix} 0 & \text{id} \\ 0 & 0 \end{pmatrix}} X_{n}\oplus X_{n+1}\to ...,$$
where $X_{n-1}\oplus X_{n}$ sits in cohomological degree $n$. 
Next, a right adjoint ${\mathcal R}$ to ${\mathcal V}$ is given by $${\mathcal R}(X) := \Sigma\ {\mathcal L}(X)\quad\equiv\quad ...\to X_n\oplus X_{n+1}\xrightarrow{\begin{pmatrix} 0 & \text{id} \\ 0 & 0 \end{pmatrix}} X_{n+1}\oplus X_{n+2}\to ...$$
where now $X_n\oplus X_{n+1}$ sits in cohomological degree $n$.
Hence we obtain the infinite chain of adjunctions
$$...\dashv {\mathcal L}\dashv {\mathcal V}\dashv {\mathcal R}\equiv\Sigma{\mathcal L}\dashv {\mathcal V}\Sigma^{-1}\dashv \Sigma {\mathcal R}\equiv\Sigma^2{\mathcal L}\dashv ...$$
Addendum The above example is rather ring-theoretical than homological: For ${\mathscr A} := k\text{-Vect}$ for a field $k$, it is essentially the observation that the ${\mathbb Z}$-graded algebra $k[x]/(x^2)$ is graded Frobenius. For an ordinary finite-dimensional algebra $A$ over $k$, being Frobenius implies that induction and coinduction from the base field are isomorphic as functors $k\text{-Vect}\to A\text{-Mod}$, but for graded Frobenius algebras this is only true up to shift by the Gorenstein parameter of the algebra - which is $1$ in the above case. Hence, graded Frobenius algebras provide many examples of infinite non-periodic chains of adjunctions, but they all have the property that they are periodic up to twisting by an automorphism.
Interestingly, the infinite version of Zhen's example from Adjoint pairs, triplets and quadruplets also exhibits this periodicity up to twisting by automorphisms: consider the poset $({\mathbb Z},\leq)$ as a category and look at the following order preserving functions / functors $({\mathbb Z},\leq)\to({\mathbb Z},\leq)$: $$\iota_n: {\mathbb Z}\to{\mathbb Z},\quad \iota_n(k) := \begin{cases} k & \text{ if } k\leq n\\ k+1 & \text{ if }k>n\end{cases},\\
\pi_n: {\mathbb Z}\to{\mathbb Z},\quad \pi_n(k) := \begin{cases} k & \text{ if } k\leq n\\ k-1 & \text{ if }k>n\end{cases}.$$
Then we have $$...\dashv \iota_n\dashv \pi_n\dashv \iota_{n-1}\dashv \pi_{n-1}\dashv ...,$$
but again $\pi_n \cong \Sigma^n \pi_0\Sigma^{-n}$ where $\Sigma : {\mathbb Z}\to {\mathbb Z}$ is the successor automorphism.
A: No, it's in fact possible for $F_1$ to be not isomorphic from $F_2$ (and this automatically implies $G_1 \not\cong G_2$, because left adjoints are unique up to isomorphism and $F_1 \dashv G_1$, $F_2 \dashv G_2$). The chain can even go on indefinitely in both directions.
Here's an example. Let $i : \mathsf{A} \to \mathsf{B}$ be a functor, and let $i^* : \mathsf{C}^\mathsf{B} \to \mathsf{C}^\mathsf{A}$ be the restriction functor. Assume everything is nice so that the Kan extensions exist. Then you have
$$\operatorname{Lan}_i \dashv i^* \dashv \operatorname{Ran}_i$$
but in general left and right Kan extensions are rarely isomorphic. By using categories of sheaves it's possible to work out an example where either $\operatorname{Ran}_i$ has a right adjoint or $\operatorname{Lan}_i$ has a left adjoint.
