= Solution
\b[The isomorphic pair is $A_8\cong PSL_4(2)$; $PSL_3(4)$ is different.] All three have order $20160$, so order alone does not settle the question; explicitly $|PSL_3(4)|=(64-1)(64-4)(64-16)/(3\cdot3)=20160$. We give a concrete realization and an element-order distinction.
On $W=\bigwedge^2\mathbb F_2^4$, the <Pfaffian> is
$$
Q(x)=x_{12}x_{34}+x_{13}x_{24}+x_{14}x_{23}.
$$
It is a nonsingular plus-type <quadratic form>: the three displayed pairs give a hyperbolic decomposition. Change of <basis> multiplies the <Pfaffian> by the <determinant>, which is always one in $GL_4(2)$. Hence the <exterior-square realization of PSL4 over F2> embeds $GL_4(2)$ into $O_6^+(2)$. The <kernel of a group homomorphism> is trivial: an element acting trivially on the <exterior square> fixes every decomposable two-space, then every one-space as an intersection of two-spaces, and hence is a scalar; over $\mathbb F_2$ this scalar is one.
For a second model let $E$ be the even-weight <vector subspace> of $\mathbb F_2^8$ and quotient by the all-one vector. The dot product descends to a <nondegenerate> <alternating bilinear form> on the six-dimensional quotient, and
$$
q(\bar x)=\operatorname{wt}(x)/2\pmod2
$$
is well defined, since complementing an even subset changes half its weight from $w/2$ to $4-w/2$. It has $(1+70+1)/2=36$ zero vectors, so is plus type. Coordinate <permutations> give a <faithful> $S_8$ action: fixing the quotient fixes each weight-two subset, because its complement has weight six, and fixing all two-subsets forces the identity <permutation>.
From the independently derived symplectic order and quadratic orbit count below,
$$
|O_6^+(2)|=\frac{|Sp_6(2)|}{36}=40320,
\qquad |GL_4(2)|=(16-1)(16-2)(16-4)(16-8)=20160.
$$
Thus $O_6^+(2)\cong S_8$ and the image of $GL_4(2)=SL_4(2)=PSL_4(2)$ is its index-two <subgroup> $A_8$. The index-two <subgroup> is unique: a nontrivial <group homomorphism> $S_8\to C_2$ sends every <transposition> to the same nonidentity element and is the sign map.
To separate $PSL_3(4)$, observe that $A_8$ contains $(12345)(678)$ of order 15. An odd-order projective element of $PSL_3(4)$ lifts to a <semisimple linear operator>, since its order upstairs divides three times an odd order in <characteristic> two. The possible <eigenspace> degree patterns are $1+1+1$, $2+1$ and $3$. In the split case all <eigenvalues> have order dividing three, so the projective element has no order 15. In the $2+1$ case the <eigenvalues> are $\lambda,\lambda^4,\lambda^{-5}$ with $\lambda\in\mathbb F_{16}^\times$; its fifth power is scalar because $\lambda^{15}=1$, so its projective order divides five. In the irreducible case the determinant-one torus has order $4^2+4+1=21$ and quotienting its scalar <subgroup> of order three gives projective order dividing seven. \b[There is therefore no element of order 15 in $PSL_3(4)$.]
Several small-parameter coincidences connect the finite <classical groups> to <permutation groups>. The projective-line action immediately gives $PSL_2(2)\cong S_3$, $PSL_2(3)\cong A_4$, $PGL_2(3)\cong S_4$ and $PSL_2(4)\cong A_5$ by <faithful> actions of degrees three, four and five and their orders. Also $PSL_2(5)\cong A_5$: its involutions have Klein-four <centralizers>, giving five <Sylow subgroups>; their <conjugation action> is <faithful> and embeds its order-60 <simple group> into $S_5$. Simplicity here can be checked from class sizes $1,15,20,12,12$: no proper nontrivial normal class union has order dividing 60. The same five-subgroup action extends to $PGL_2(5)$, with trivial <kernel of a group homomorphism> and order 120, giving $S_5$. Indeed its kernel intersects $PSL_2(5)$ trivially, so centralizes that normal <subgroup>. Commuting projectively with both upper and lower elementary unipotents forces a representing matrix to be scalar, so this kernel is trivial.
The less immediate coincidence $PSL_2(9)\cong A_6$ can be exhibited on the ten unordered triple partitions used in Question 4. Let $a=(123)$, $b=(456)$, $w=(12)(36)$, assign $\infty$ to $123\mid456$ and $0$ to $126\mid345$, and label the other partitions by applying $a^xb^y$ to the zero partition, with $t=x+iy\in\mathbb F_9$ and $i^2=-1$. Inspection of these nine partitions gives
$$
a:t\mapsto t+1,\qquad b:t\mapsto t+i,\qquad w:t\mapsto-1/t.
$$
Translations and inversion generate $PSL_2(9)$: conjugating the upper unitriangular matrices by inversion gives the lower ones, and elementary elimination generates $SL_2(9)$. Their projective action has order $9(9^2-1)/2=360$. These partition <permutations> come from even <permutations> of six letters, whose partition action is <faithful> as proved below; hence they fill $A_6$.
Finally, on the even-weight module of $\mathbb F_2^6$ modulo its all-one vector, coordinate <permutations> preserve a <nondegenerate> <alternating bilinear form> in <dimension> four and act faithfully, again by the weight-two-subset argument. Both $S_6$ and $Sp_4(2)$ have order 720, proving $Sp_4(2)\cong S_6$, with <commutator subgroup> $A_6$. These examples illustrate both the need to quotient <group centers> and the low-rank exceptions to general simplicity. They do not make equal orders sufficient for <group isomorphism>, as the two order-20160 examples already demonstrate.
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