Your property is too strong; there is no Suslin line like that. 

The reason is that every Suslin line contains a copy of the real line, and we can define a counterexample function $f$ that concentrates only on this copy of $\mathbb{R}$. Specifically, suppose that $L$ is a Suslin line. Since the order is dense, we may find a copy of the rational line $\mathbb{Q}$ inside $L$, that is, a suborder $Q\subset L$ which is a countable dense endless order. Since $L$ is complete, we may add points to $Q$ realizing all the Dedekind cuts in $Q$, thereby extending $Q$ to a suborder $R\subset L$ that is order-isomorphic to the real line $\mathbb{R}$. Now, we may define $f$ on this copy of $\mathbb{R}$ to be the analogue of adding one, say, or any other order-preserving map with no fixed points. This gives an injective order-preserving map $f:R\to R$ for an uncountable subset $R\subset L$, with no fixed points. 

Meanwhile, a positive answer is possible if one considers an analogue of your property on the underlying Suslin tree. I shall write this up and make an edit later on.