I claim that if $X$ is a linear ordering and $f:X\rightarrow X$ is an order preserving surjective mapping which is not the identity mapping, then there is an order preserving injective mapping $g:X\rightarrow X$ which is not the identity mapping.

If $f$ is surjective but not the identity function, the there is some function $g:X\rightarrow X$ such that $f\circ g$ is the identity mapping. The mapping $g$ is clearly injective and not the identity mapping. I now claim that $g$ is order preserving. Suppose that $x<y$. Then $f(g(x))=x<y=f(g(y))$. Therefore, since $f$ is order preserving, we have $g(x)<g(y)$ since if $g(x)\geq g(y)$, then $f(g(x))\geq f(g(y))$ which is a contradiction. Therefore, $g$ is our desired injective order preserving mapping that is not the identity mapping.

On the other hand, by this answer, if $2^{<\kappa}=\kappa$, then there is a linear order $X$ of cardinality $2^{\kappa}$ with no order preserving injections $f:X\rightarrow X$. Furthermore, under GCH, for every uncountable cardinal $\kappa$ there is a total order $X$ with $|X|=\kappa$ and where there is no order preserving injective function from $X$ to $X$. As a consequence, if $2^{<\kappa}=\kappa$, then there is a linear order $X$ of cardinality $2^{\kappa}$ with no order preserving surjections $f:X\rightarrow X$, and
under GCH for each uncountable cardinal $\kappa$ there is a total order $X$ with $|X|=\kappa$ and where there is no order preserving surjective mapping $f:X\rightarrow X$.