A relation on a nonempty set cannot be both reflexive and irreflexive. This follows almost immediately from the definitions: a reflexive relation on a nonempty set $X$ must contain every pair of the form $(x,x) \in X\times X$, while an irreflexive relation cannot contain any such pair. Reflexivity and irreflexivity are mutually exclusive properties.
Aside from such vacuous examples (vacuous in the sense that the hypotheses are false, not in the sense that they are "easy"), a relation cannot be both transitive and intransitive: if $(x,y), (y,z) \in R$, then either $(x,z) \in R$ (and $R$ is not intransitive), or $(x,z) \not\in R$ (and $R$ is not transitive). Aside from vacuous examples, these two properties are mutually exclusive.
A nontrivial relation which is intransitive must also be irreflexive. The essential idea here is that reflexive relations "build in" transitive relations. More formally, consider a proof by contraposition: suppose that $R$ is a nontrivial relation which is not irreflexive. Then there is some $x$ such that $(x,x) \in R$. Taking $x=y=z$, this implies that
$$ (x,y), (y,z), (x,z) \in R, $$
which contradicts the definition of intransitivity. Thus $R$ is not intransitive. Therefore a relation which is not irreflexive is not intransitive.
By contraposition, an intransitive relation must be irreflexive.
Perhaps counterintuitively, a nontrivial relation can be both symmetric and antisymmetric. Suppose that $R$ is a nontrivial relation which is both symmetric and antisymmetric. As $R$ is nontrivial, it contains some pair $(x,y)$. The symmetry of $R$ implies that $(y,x)$ is also in $R$. The antisymmetry of $R$ then implies that $x=y$. Hence a relation on a set $X$ which is both symmetric and antisymmetric must be a subset of the diagonal $\{(x,x) : x \in X\}$. Any such relation is vacuously transitive, and can be reflexive if it is the entire diagonal (this is the equality relation). There is no nontrivial irreflexive relation which is both symmetric and antisymmetric.
Commentary is given in cases where it might be illuminating.
[RTS] $R = \{(1,1), (1,2), (2,2), (2,1), (3,3)\}$
[RTA] $R = \{(1,1), (1,2), (1,3), (2,2), (2,3), (3,3)\}$
[RT-] $R = \{ (1,1), (1,2), (2,1), (2,2), (3,1), (3,2), (3,3) \}$
[RIS] No example exists, see 3.
[RIA] No example exists, see 3.
[RI-] No example exists, see 3.
[R-S] $R = \{(1,1), (1,2), (2,1), (2,2), (2,3), (3,2), (3,3)\}$
[R-A] $R = \{(1,1), (1,2), (2,2), (2,3), (3,3)\}$
[R--] $R = \{(1,1), (2,2), (3,3), (1,2), (2,3) \}$
[ITS] No nontrivial example exists.
Suppose that $R$ is some nontrivial, irreflexive, transitive relation. If $R$ is not antisymmetric, then there exist pairs $(x,y)$ and $(y,x)$ which are both elements of $R$. But $R$ is transitive, so $(x,x)$ and $(y,y)$ must also be elements of $R$. In other words, a nontrivial, irreflexive, transitive relation must be antisymmetric.
[ITA] $R = \{(1,2), (1,3), (2,3)\}$
The usual order relations ($\le$, $<$, $\ge$, $>$) on $\mathbb{R}$ are more interesting examples of relations which are transitive and antisymmetric. Weak inequalities are reflexive, while strict inequalities are irreflexive.
[IT-] No nontrivial example exists, see [ITS].
[IIS] $\{(1,2), (2,1)\}$.
[IIA] $R = \{(1,2)\}$
[II-] $\{(1,2), (1,3), (2,1)\}$.
[I-S] $R = \{(1,2), (2,1), (2,3), (3,2) \}$.
Transitivity and intransitivity can be a little hard to see by inspection. This relation is not intransitive, as every intransitive relation must be antisymmetric; and it is not transitive, as $(1,2),(2,3) \in R$ but $(1,3)\not\in R$.
There is no example of an irreflexive and antisymmetric relation on $X$ which is neither transitive nor intransitive. However, if $R$ is a relation on as set $Y = \{a,b,c,d\}$, then an example exists:
[I-A] $R = \{ (a,b), (a,c), (b,c), (c,d) \}$
This relation is not transitive, because $(a,c), (c,d) \in R$, but $(a,d)\not\in R$; and is not intransitive, because $(a,b), (b,c), (a,c) \in R$.
[I--] $R = \{(1,2), (1,3), (2,1), (2,3)\}$
[-TS] $R = \{(1,1), (1,2), (2,1), (2,2)\}$
Note that the above relation is not reflexive on the three element set $X = \{1,2,3\}$ because it does not contain the pair $(3,3)$. However, thought of as a relation on the two element set $\{1,2\}$, this relation is reflexive.
[-TA] $R = \{(1,1), (1,2), (2,3), (3,1)\}$
[-T-] $R = \{(1,1), (1,2), (1,3), (2,1), (2,2), (2,3) \}$
[-IS] No nontrivial example exists, see 3.
[-IA] No nontrivial example exists, see 3.
[-I-] No nontrivial example exists, see 3.
[--S] $R = \{(1,2), (2,1), (2,2) \}$
[--A] $R = \{(1,1), (1,2), (2,3) \}$
[---] $R = \{ (1,1), (1,2), (2,1), (2,3) \}$
Abstractly, it is good to have simple examples and counterexamples to different permutations of relational properties. However, it is also useful to have in mind more interesting models—every one of these properties comes from something in the world. The arbitrary permutations of properties may not have any useful meaning, but the properties themselves are interesting.
An equivalence relation is any relation which is reflexive, transitive, and symmetric. The most basic such relation is equality ($=$): $x=y$ if and only if $x$ and $y$ are, in fact, the same object. Abusing notation a bit, this means that $=$, though of as an equivalence relation on some arbitrary set $X$, is the diagonal of $X\times X$. That is,
$$ = \quad\text{is the set}\quad \{ (x,x) : x \in X\}. $$
There are other important equivalence relations, and many important properties in mathematics hold only "up to equivalence" with respect to some equivalence relation.
For example, $1/2$ and $2/4$ are not really the same object—ask any second grader. If I have a package of two cookies, then I can have one cookie, and give another to a friend. We each get one of the two cookies, or $1/2$ of the package. If I have a package of four cookies, then I can have two and give two to a friend. We each get two cookies, or $2/4$ of the package. Two is not one! These things are different. However, from the point of view of addition and multiplication, $1/2$ and $2/4$ behave in essentially the same way—they are equivalent with respect to a relation which ultimately gives us the rational numbers. Hence we can treat them as though they are the same object (and typically do!).
Order relations are examples of transitive, antisymmetric relations. For example, $\le$, $\ge$, $<$, and $>$ are examples of order relations on $\mathbb{R}$—the first two are reflexive, while the latter two are irreflexive. Set containment relations ($\subseteq$, $\supseteq$, $\subset$, $\supset$) have simililar properties.
In general, I think that it is reasonable to think of transitive, antisymmetric relations as those relations which "rank" or "order" things in some rough way. Inequalities order numbers, set containment relations order sets, taxonomies classify and order living organisms, etc.
Intransitive relations are kind of an odd duck, and it is not immediately obvious how they might come up in the real world. However, they do! My favorite example is the two-player game "Rock-Paper-Scissors". Rock beats scissors, scissors beats paper, paper beats rock. The relation "beats" is intransitive. Parenthood is also (generally speaking—one can always find exceptions once human behaviour is involved) an intransitive relation: I am the parent of my daughter, and my mother is my parent, but my mother is not my daughter's parent.
