Here is an explicit isomorphism between the two groups using an intermediate group:
$$\begin{array}{|l|l|l|}
\hline
PSL(2,\mathbb{Z_3})& \ \ \ Homog(2,\mathbb{Z_3})& \ \ \ \ \ \ \ \ \ \ \ \ \ A_4\\
\hline
\pmatrix{1&0\\0&1}&h_{1}(z)=z&\pmatrix{0 & 1 & 2 & \infty\\0 & 1 & 2 & \infty}\\ \hline
\pmatrix{1&0\\1&1}&h_{2}(z)=\dfrac{z}{z+1}&\pmatrix{0 & 1 & 2 & \infty\\0 & 2 & \infty & 1}\\
\hline
\pmatrix{1&1\\0&1}&h_{3}(z)=z+1&\pmatrix{0 & 1 & 2 & \infty\\1 & 2 & 0 & \infty}\\
\hline
\pmatrix{1&0\\2&1}& h_{4}(z)=\dfrac{z}{2z+1}&\pmatrix{0 & 1 & 2 & \infty\\0 & \infty & 1 & 2}\\
\hline
\pmatrix{1&1\\1&2}&h_{5}(z)=\dfrac{z+1}{z+2}&\pmatrix{0 & 1 & 2 & \infty\\ 2 & \infty & 0 & 1}\\
\hline
\pmatrix{1&1\\2 & 0}&h_{6}(z)=\dfrac{z+1}{2z}&\pmatrix{0 & 1 & 2 & \infty\\ \infty & 1 & 0 & 2}\\
\hline
\pmatrix{1&2\\0 & 1}&h_{7}(z)=z+2&\pmatrix{0 & 1 & 2 & \infty\\ 2 & 0 & 1 & \infty}\\
\hline
\pmatrix{1&2\\1 & 0}&h_{8}(z)=\dfrac{z+2}{z}&\pmatrix{0 & 1 & 2 & \infty\\ \infty & 0 & 2 & 1}\\
\hline
\pmatrix{2&1\\1 & 1}&h_{9}(z)=\dfrac{2z+1}{z+1}&\pmatrix{0 & 1 & 2 & \infty\\ 1 & 0 & \infty & 2}\\
\hline
\pmatrix{0&2\\1 & 0}&h_{10}(z)=\dfrac{2}{z}&\pmatrix{0 & 1 & 2 & \infty\\ \infty & 2 & 1 & 0}\\
\hline
\pmatrix{0&2\\1 & 1}&h_{11}(z)=\dfrac{2}{z+1}&\pmatrix{0 & 1 & 2 & \infty\\ 2 & 1 & \infty & 0}\\
\hline
\pmatrix{0&1\\2 & 1}&h_{12}(z)=\dfrac{1}{2z+1}&\pmatrix{0 & 1 & 2 & \infty\\ 1 & \infty & 2 & 0}\\
\hline
\end{array}$$
i.e., we are going to establish a combined isomorphism between 3 groups, the middle one making the connection between the two extreme ones:
The group $PSL(2,\mathbb{Z_3})$ for matrix multiplication,
The group of homographic functions on $\mathbb{Z_3}$ extended to $\mathbb{Z_3} \cup \{ \infty\}$ for the composition of functions (see below), and
The group $A_4$ of even permutations on 4 objects $\mathbb{Z_3} \cup \{\infty\}$, for the composition of bijections.
This "Rosetta stone" deserves of course a detailed explanation.
Let us begin by recalling what is $PSL(2,\mathbb{Z_3})$ by using a kind of etymological analysis:
(1) P is as "Projective", [see below],
(2) SL2 as "Special Linear", $2$-dimensional, more precisely: "L2" is for $2 \times 2$ matrices and "S" (special) for determinant equal to 1 (mod. 3),
(3) the entries of these matrices being in $\mathbb{Z_3}=\{\bar{0},\bar{1},\bar{2}\}$ with the usual addition and multiplication mod. 3, with the following rules: $3 \equiv 0$, $2 \equiv -1$, $2 \times 2 \equiv 1.$
About the term "projective": two matrices are in the same equivalence class (thus are considered as a same projective "object") iff they obey the equivalence rule:
$$\tag{*}\exists k \neq 0 \ \ \text{such that} \ \ \pmatrix{a&c\\b&d}=k \pmatrix{a'&c'\\b'&d'}.$$
In fact, the unique non trivial case is with $k \equiv 2$ (mod $3$).
Therefore, every equivalence class, i.e., every coset, has 2 elements.
For example, the two following matrices are considered as a same element:
$$\pmatrix{2&0\\1&2} \sim \pmatrix{1&0\\2&1},$$
because the first one is twice the second and reciprocally (only one of them is present in the first column of the table upwards).
Remark: The equivalence relationship (*) is the same as yours because the center of group $GL(2,\mathbb{Z_3})$ is the group with two elements:
$$I=\pmatrix{1&0\\0&1} \ \ \text{and} \ \ 2 I=\pmatrix{2&0\\0&2}$$
constituting the neutral class (the "coset" of $I$).
It's time now to explain the isomorphism between $PGL(2,\mathbb{Z_3})$ and $A_4$, that has been explicited in the table at the beginning.
Every matrix $M=\pmatrix{a&c\\b&d} \in PGL(2,\mathbb{Z_3})$, can be associated with a unique "homography":
$$z \rightarrow h_M(z)=\frac{az+c}{bz+d}$$
Why this uniqueness ? Because equivalence relationship (*) "up to a non-zero multiplicative factor $k$" is associated with a cancellation when a same factor $k$ is present in the numerator and in the denominator of a fractional expression.
Let us now consider the images of $(0,1,2,\infty)$ by $h_M.$
Remark: the presence of $\infty$ isn't a true surprise in such a context because, in general, projective geometry uses infinity almost constantly. Without giving a fully rigorous explanation, it suffices to say that the rules governing the special symbol $\infty$ are the same as the rules dealing with limits "when the variable tends to $\infty$" in real analysis.
Let us consider the example of $M=\pmatrix{1&1\\1&2}$, and its associated homography $h(z)=\frac{1z+1}{1z+2}=1-\frac{1}{z+2}$. Using the upsaid rules, we have:
$$\begin{cases}h(0)&=&1/2=1.2^{-1}=2,\\h(1)&=&"1/0"=\infty,\\h(2)&=&0/4=0,\\h(\infty)&=&1 \ - "1/\infty"=1-0=1. \end{cases}$$
and the fact is that this permutation $\pmatrix{0 & 1 & 2 & \infty\\2 & \infty & 0 & 1}$ on the 4 objects $(0,1,2,\infty)$ is even...
A "rationale" for this evenness : The only possible subgroup with 12 elements of the symmetrical group with $4!=24$ elements is the group of even permutations $A_4$ ($A_n$ is the only subgroup of $S_n$ of index $2$.).
Remark : for projective geometry and homographies on finite fields or rings, see (https://en.wikipedia.org/wiki/Projective_line_over_a_ring).
Edit: using SAGE (calling sagecell.sagemath.org), one can have further informations, for example :
G=PSL(2,3);G
Answer : Permutation Group with generators [(2,3,4), (1,2)(3,4)]
With the request :
G.list()
one gets the representation
[(),
(1,3)(2,4),
(1,4)(2,3),
(1,2)(3,4),
(2,4,3),
(1,3,4),
(1,4,2),
(1,2,3),
(2,3,4),
(1,3,2),
(1,4,3),
(1,2,4)]
(under the form of permutations, presented as composition of cycles). A good and simple exercise is to establish a correspondence between letters $a,b,\cdots l$ and the last column of the array above. The following request :
G.cayley_table()
gives :
* a b c d e f g h i j k l
+------------------------
a| a b c d e f g h i j k l
b| b c a f d e h i g l j k
c| c a b e f d i g h k l j
d| d g j a h k b e l c f i
e| e i k c g l a f j b d h
f| f h l b i j c d k a e g
g| g j d k a h e l b i c f
h| h l f j b i d k c g a e
i| i k e l c g f j a h b d
j| j d g h k a l b e f i c
k| k e i g l c j a f d h b
l| l f h i j b k c d e g a
If furthermore, you ask :
G.is_isomorphic(AlternatingGroup(4))
the answer is :
True...
