Here is a simple example of size continuum. Do the ordinary middle-third construction of the Cantor set, except that whenever you delete the $n$-th (numbered by level and then left to right, say) middle-third interval leave in exactly $n$ points from that interval. Let's call the resulting set $X$. Any automorphism of $X$ must map (resp. left-, right-) isolated points to (resp. left-, right-) isolated points. Since we left a different number of points in each middle-third interval, the automorphism must fix every inner point of each middle-third interval as well as its two endpoints. By density, it follows that the automorphism fixes every point of $X$.
All of your examples are scattered, I checked the Rosenstein's Linear Orderings to see if he had anything good to say about which scattered linear orders are rigid. To my dismay, this is what I found: "These considerations seem to make impossible an inductive argument (on $F$-rank or $VD$-rank) to determine which scattered types are rigid." (p. 133) However, he does cite a result of Anne Morel (Ordering relations admitting automorphisms, Fund. Math. 54 (1964), 279-284.) which says that a linear order $A$ is not rigid if and only if $A cong A_1 + A_2timesmathbb{Z} + A_3$ for some linear orderings $A_1,A_2,A_3$, with $A_2$ nonempty.
Dense examples of rigid subsets of $mathbb{R}$ would be interesting to see. This is probably not too difficult to construct under CH. But a ZFC example might have to deal with important barriers such as Baumgartner's result that All ${aleph_1}$-dense subsets of $mathbb{R}$ can be isomorphic, Fund. Math. 79 (1973), 101-106. Maybe there are some examples in this classic paper of Sierpinski Sur les types d'ordre des ensembles linéaires, Fund. Math. 37 (1950), 253-264.
All three papers can be found here.
Addendum (after sdcvvc's comment): For the sake of completeness, I'm including a simplification of the Dushnik-Miller argument that produces a dense subset $X$ of $mathbb{R}$ which is rigid (though not the stronger result that $X$ has no self-embeddings).
To ensure density, the set $X$ will contain all rational numbers. Note that an automorphism $f$ of $X$ is then completely determined by its restriction to $mathbb{Q}$. Indeed, since $f[mathbb{Q}]$ must be dense (in $X$ and) in $mathbb{R}$, we always have
$f(x) = sup{f(q):q in (-infty,x)capmathbb{Q}} = inf{f(q):q in (x,infty)capmathbb{Q}}.$
There are only $c = 2^{aleph_0}$ increasing maps $f:mathbb{Q}tomathbb{R}$ with dense range. Let $langle f_alpha:alpha<c rangle$ enumerate all such maps, except for the identity on $mathbb{Q}$.
We will define by induction a sequence $langle (x_alpha,y_alpha) : alpha<c rangle$ of pairs of irrational numbers. The $x_alpha$ will be points of $X$ while the $y_alpha$ will be in the complement of $X$. For each $alpha$, we will have $f_alpha(x_alpha) = y_alpha$ (in the sense of the inf/sup definition above).
Suppose we have defined $(x_beta,y_beta)$ for $beta<alpha$. Since $f_alpha$ is not the identity, there is a rational $q$ such that $f_alpha(q) neq q$. Let's suppose that $f_alpha(q) > q$ (the case $f_alpha(q) < q$ is symmetric). Since the real interval $(q,f_alpha(q))$ has size $c$ and the extension of $f_alpha$ to all of $mathbb{R}$ is injective, we can always pick
$x_alpha in (q,f_alpha(q)) setminus(mathbb{Q}cup{y_beta:beta<alpha})$
such that $y_alpha = f_alpha(x_alpha) notin mathbb{Q}cup{x_beta:beta<alpha}$. Note that $x_alpha < f_alpha(q) < y_alpha$ so $x_alpha neq y_alpha$.
In the end, we will have
${x_alpha: alpha < c} cap {y_alpha : alpha<c} = varnothing$
and any set $X$ such that
$mathbb{Q}cup{x_alpha:alpha<c} subseteq X subseteq mathbb{R}setminus{y_alpha:alpha<c}$
is necessarily rigid since $f_alpha(x_alpha) = y_alpha notin X$ for each $alpha<c$.
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