If the $Aut(C)$ is trivial for a chain $C$ then $C$ or $C^{op}$ is a well order? If you have a chain that is a well order (all well orders are chains), then $Aut(C)$ is trivial. I'm asking if the converse is also true.
 A: The converse is false.  Here's a somewhat ridiculous way to construct a counterexample.  We construct $C$ as the union of an increasing sequence of subchains $C_0\subset C_1\subset C_2\subset\dots$.  Start by letting $C_0$ be your favorite chain that is neither well-ordered nor co-well-ordered (say, $C_0=\mathbb{Q}$).  Now given $C_n$, we construct $C_{n+1}$ as follows.  For each element $x\in C_n\setminus C_{n-1}$, choose a different regular cardinal $\kappa_x$ which is greater than $|C_n|$.  We then add to $C_n$ an increasing sequence of length $\kappa_x$ which is less than $x$ but greater than every element of $C_n$ which is less than $x$.  Let $C_{n+1}$ be the chain obtained by adding these sequences for all $x\in C_n\setminus C_{n-1}$.
Finally, we let $C=\bigcup_n C_n$.  Since $C_0\subset C$, $C$ is neither well-ordered nor co-well-ordered.  But I claim $C$ has no nontrivial automorphisms.  Indeed, for any $x\in C$, the cofinality of $x$ from below is $\kappa_x$, the cardinal we chose when constructing $C_{n+1}$ (if $x\in C_n\setminus C_{n-1}$).  These cardinals are different for all different values of $x$, so every element of $C$ has a different cofinality from below.  Thus an automorphism of $C$ can only send $x$ to itself for each $x$, and so must be the identity.

Somewhat less ridiculously, you can find a counterexample that is a subset of $\mathbb{R}$ by transfinite induction.  Here's a sketch of the construction.  Start by declaring that $\mathbb{Q}\subseteq C$.  Note that then that $\mathbb{Q}$ will be dense in $C$, so any automorphism of $C$ is uniquely determined by its restriction to $\mathbb{Q}$.  There are only $2^{\aleph_0}$ maps $\mathbb{Q}\to\mathbb{R}$, so you can diagonalize them: one by one, decide whether elements of $\mathbb{R}$ are either in $C$ or not in $C$, in order to make each non-identity map $\mathbb{Q}\to\mathbb{R}$ fail to extend to an automorphism of $C$. 
