An automorphism is an isomorphism of an algebraic structure with itself.
Let's review an isomorphism $i$ of two linearly ordered groups $G(+,<)$ and $F(*,<)$.
$i$ is linear order-preserving when $a < b \iff i(a) < i(b)$ for any elements $a$ and $b$ in $G$.
Let's note that every linear order induces the reverse order: $G(+, <, >)$, $F(*, <, >)$.
An isomorphism $i^*$ is linear order-reversing when $a < b \iff i(a) > i(b) \iff i(b) < i(a)$.
Any linear order-preserving isomorphism $i$ induces linear order-reversing isomorphism $i^*$.
In case $G$ and $F$ are different, choosing between $<$ and $>$ in each of the groups is an arbitrary decision, so one can say $i$ and $i^*$ are equivalent.
It looks like this is not the case when we talk about automorphisms.
Let's assume $G$ is commutative and review the inversion automorphism $i(a) = -a$.
$i$ is order-reversing since $a < b \iff -b < -a$.
But in this case we can clearly see the difference between order-preserving and order-reversing automorphisms: $i$ maps $<$ to $>$ on the same set.
I am wondering if it make sense to explicitly say if an automorphism keeps operations and relations the same?
I've never thought of mapping $+$ to $\times$ in an automorphism of an algebraic structure $(+, \times)$.
Are there practical examples of such mapping?
There was a similar discussion with no positive conclusion in here: https://www.researchgate.net/post/Difference_between_an_order_automorphism_and_order_isomoprhism