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.

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. Your property on the underlying Suslin trees is a kind of strong rigidity property of the kind considered in my article

• G. Fuchs and J. D. Hamkins, Degrees of rigidity for Souslin trees, J. Symbolic Logic, vol. 74, iss. 2, pp. 423-454, 2009.

In that article, we define that a Suslin tree $T$ has the unique branch property if forcing with $T$ necessarily adds only one branch through the tree. This is a strengthening of rigidity, since if $T$ has a nontrivial automorphism $\pi$, then one find a condition $x$ in $T$ with $\pi(x)\neq x$, and then forcing below $x$ will add two branches, namely, the generic branch $g$ and also $\pi[g]$, which will be different. We show that the Suslin trees obtained by the usual forcing to add a Suslin tree have the unique branch property, and also one can construct them in $L$ using the $\Diamond$ principle.

This is an analogue of your property for the trees:

Theorem. If $T$ is a Suslin tree with the unique branch property, then there is no function $f:S\to T$ with $S$ an uncountable set of extensions of a fixed node $x$, where $f(x)\perp x$.

Proof. Consider the uncountable set $S\subset T$. We may regard $S$ as a tree on its own, and in fact, $S$ is a Suslin tree. Force with $S$, so that we add an $\omega_1$-branch $b$ through $s$ in $V[b]$. It follows that the image $f[b]$ of this branch is a cofinal branch through $T$, and this is a different branch because $x\perp f(x)$. But note that $b$ is a cofinal branch through $T$, and since $T$ is Suslin, it follows that $b$ is also generic for forcing over $T$. Thus, in the forcing extension $V[b]$, we have two branches through $T$, which violates the unique branch property. QED