A well-known fact is that every automorphism of the symmetric group on a set must be inner (whether the set is finite or infinite) unless the set has exactly six elements (4, § 13). A long-standing conjecture concerns the analogue of this fact for the group A(S) of all order-preserving permutations of a totally ordered set S. The group A(S) is lattice ordered (l-group) by defining, for f, g ∈ A (S), f ≦ g whenever xy ≦ xg for all x ∈ S. From the standpoint of l-groups, A(S) is of considerable interest because of the analogue of Cayley's theorem proved in (2), namely every l-group may be embedded in some A(S). Unlike the non-ordered symmetric groups, which must be highly transitive, A(S) is severely restricted if assumed transitive. Of special interest are those cases when A(S) is doubly transitive (relative to the order), since the building blocks (primitive) of all transitive A(S) are either doubly transitive or uniquely transitive. It is easily seen that each inner automorphism of A(S) must preserve the lattice ordering (that is, must be an l-automorphism). It is tempting to conjecture that every l-automorphism of A(S) must be inner. However, an easy counterexample is at hand. If S consists of two copies of the ordered set of integers, one entirely above the other, then A(S) is the direct product of two copies of the l-group of integers, which has an outer l-automorphism exchanging the two factors. The conjecture referred to above is that if A(S) is transitive on S, then every l-automorphism of A(S) is inner.