a)The action that is transitive and faithful
Your Answer: Group G under addition acting on a set of integers Z
It's not an answer because you did not say how it is acting (see also Leon Aragones'comment). By the way, $\mathbb{Z}$ is always acting in a natural way on $G$ when $G$ is abelian (but the action might not be transitive).
My Hint (trivial) : The trivial group $G$ acts on $\{1\}$ trivially, this should do the job why ?
My Hint (a little less trivial) : Take $G$ a group acting on itself in some particular way...
b)The action that is transitive but is NOT faithful
Your Answer: Group G of 60 degree rotations acting on a set of Vertices of Dihedral group $D_3$ since all 3 rotations fix everything.
"set of Vertices of Dihedral group $D_3$" do you mean the set of vertices of the equilateral triangle $T$ for which $D_3=Isom(T)$? In that case there are 2 rotations which does not fix everything.
My Hint (trivial) : You should go more simple, take $X:=\{1\}$ (i.e. a set with one element) then a group $G$ always acts on $X$ (the action is unique why?). Under which conditions on $G$ is the action transitive? faithful?
My Hint (a little less trivial) : Take $G$ be a group with at least one proper subgroup $H$ which is not trivial then $G$ acts naturally on $G/H$ (left $H$-cosets). This should do the job.
c)The action that is NOT transitive but IS faithful
Can I simply say it's just a group of symmetries of a single point? Since it only has 1 element it can't be transitive, right?
It seems about right but you do need to give the whole set up.
My Hint (trivial) : Take $G$ to be the trivial group and $X$ be any set. Then there is only one action of $G$ on $X$. Under which condition on $X$ is the action transitive? faithful?
My Hint (a little less trivial) : Take $G:=GL_2(\mathbb{R})$, I claim that there is a natural action on $\mathbb{R}^2$, this one do the job.
d) The action that is NOT transitive and NOT faithful
Something that is non abelian? I don't really know.
Go simple.
My Hint (trivial) : take $G$ a group and $X$ a set. Define the action of $G$ on $X$ by $g.x:=x$. Under which conditions on $G$ and $X$ is the action transitive? faithful?
My Hint (a little less trivial) : Take $G$ be a group with at least one proper subgroup $H$ which is not trivial then $G$ acts naturally on $G/H$ (left $H$-cosets). Then $G$ also acts diagonally on $G/H\times G/H$. This should do the job.
e) The action with 2 orbits
A line with vertices 1 and 2. - The group of symmetries acting on a set of vertices.
In that case the action has only one orbit, hasn't it?
My Hint (trivial) : Trivial group and a set with two elements.
My Hint (a little less trivial) : Think about a square and an axial symmetry.
f) The action with 3 orbits
I have a triangle in my mind, but rotational symmetries of a triangle are stabilizers aren't they? Same goes for part (e).
My Hint (trivial) : Trivial group and a set with three elements.
My Hint (a little less trivial): Think about an hexagon and an axial symmetry.
Best Answer
Let us remember the definition of faithful, transitive and free:
The action of a group $G$ on a set $X$ is called:
Transitive: if $X$ is non-empty and if for any $x,y\in X$ there exists $g\in G$ such that $g\cdot x = y$.
Faithful (or effective): if for any two distinct $g,h\in G$ there exists $x\in X$ such that $g\cdot x \neq h\cdot x$; or equivalently, if for any $g\neq e\in G$ there exists $x\in X$ such that $g\cdot x \neq x$. Intuitively, in a faithful group action, different elements of $G$ induce different permutations of $X$.
Free: if, given $g,h\in G$, the existence of $x\in X$ with $g\cdot x = h\cdot x$ implies $g = h$. Equivalently: if $g$ is a group element and there exists an $x\in X$ with $g\cdot x = x$ (that is, if $g$ has at least one fixed point), then $g$ is the identity.
Now we can determine your three group actions:
The action of $S_n$ on a set of $n$-elements is as you showed transitive. It is also faithful as you explained, but it is not free, because there are elements on $S_n$ that do have fixed points that are not the identity, for example $1\to 1, 2\to 3, 3\to 4, n-1\to n, n\to 2$. Since the action is transitive the group of orbits is the group of one element.
The action of $O(n)$ on $S^{n-1}$, it is transitive, since there is always a rotation that can send any point of $S^{n-1}$ onto the north pole. For $n>2$, It is not free since for any point on $S^{n-1}$ there are many rotations different from the identity, whose axis of rotation is the line that passes through that point and the origin, and all of these fix that point. It is not faithful neither since any element of $O(n)$ that is not the identity moves $S^{n-1}$, therefore there exists a point on $S^{n-1}$ that is not fixed by the action. Since the action is transitive the group of orbits is the group of one element. The case $n=2$ the action it is transitive, free but not faithful.
The action of $O(n)$ on $\mathbb{R}^n$ is not free since every element of $O(n)$ fixes the origin but is not necessarily the identity. It is not transitive either, since for two elements of $\mathbb{R}^n$ whose length are not equal, you cannot find an element of $O(n)$ that sends one on the other (the elements of $O(n)$ preserve norm). It is not faithful neither since when you restrict the action to $S^{n-1}$ it is not faithful either. Since $\mathbb{R}^n=[0,\infty)\times S^{n-1}$ the set of orbits can be identified with $[0,\infty)$ since when $O(n)$ is restricted to the second factor it is transitive.