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"Each automorphism $\sigma$ of $G$ can be written $\sigma = g f d i$, with $i$, $d$, $f$, and $g$ being inner, diagonal, field, and graph automorphisms, respectively" (Steinberg - Automorphisms of finite linear groups, 3.2). Here, as best as I can tell, $G$ is $\operatorname{PSL}(2, \mathbb F_q)$ (not $\operatorname{PGL}(2, \mathbb F_q)$); the definition of $G$ relies on a set that looks like $\mathfrak B$, whose definition I cannot find. A diagonal automorphism is one that arises by conjugation in the diagonal subgroup of $\operatorname{PGL}(2, \mathbb F_q)$, not just of $\operatorname{PSL}(2, \mathbb F_q)$; the only non-inner automorphism that we get this way is conjugation by $\begin{pmatrix} \epsilon & 0 \\ 0 & 1 \end{pmatrix}$, where $\epsilon$ is a non-square in $\mathbb F_q^\times$.

If an automorphism $\sigma$ of $\operatorname{SL}(2, \mathbb F_q)$ induces the identity on $\operatorname{PSL}(2, \mathbb F_q)$, then $g \mapsto g^{-1}\sigma(g)$ maps $\operatorname{SL}(2, \mathbb F_q) \to \{\pm1\}$. However, $\operatorname{SL}(2, \mathbb F_q)$ is generated by its unipotent elements, hence admits no non-trivial homomorphisms to a 2-torsion group (assuming $p \ne 2$). (EDIT: If $p = 2$, then $\operatorname{SL}(2, \mathbb F_q) = \operatorname{PSL}(2, \mathbb F_q)$, so there is no ambiguity.)

EDIT: I somehow misread the question as asking about the full automorphism group of $\operatorname{SL}(2, \mathbb F_q)$, not just the outer automorphism group; and it's always a good time to break out a result of Steinberg. As @RichardLyons and @YCor point out, the outer automorphism group (of $\operatorname{SL}(2, \mathbb F_q)$, which, we argued above, is the same as that of $\operatorname{PSL}(2, \mathbb F_q)$) is $\langle\operatorname{Fr}\rangle \times \langle\operatorname{Int}\begin{pmatrix} \epsilon & 0 \\ 0 & 1 \end{pmatrix}\rangle \cong \operatorname C_n \times \operatorname C_2$, generated by the Frobenius and an appropriate conjugation in $\operatorname{PGL}(2, \mathbb F_q)$.

If $p = 2$, then, again as @YCor points out there is no extra conjugation coming from $\operatorname{PGL}(2, \mathbb F_q)$, so we just get the outer automorphism group $\langle\operatorname{Fr}\rangle \cong \operatorname C_n$.

"Each automorphism $\sigma$ of $G$ can be written $\sigma = g f d i$, with $i$, $d$, $f$, and $g$ being inner, diagonal, field, and graph automorphisms, respectively" (Steinberg - Automorphisms of finite linear groups, 3.2). Here, as best as I can tell, $G$ is $\operatorname{PSL}(2, \mathbb F_q)$ (not $\operatorname{PGL}(2, \mathbb F_q)$); the definition of $G$ relies on a set that looks like $\mathfrak B$, whose definition I cannot find. A diagonal automorphism is one that arises by conjugation in the diagonal subgroup of $\operatorname{PGL}(2, \mathbb F_q)$, not just of $\operatorname{PSL}(2, \mathbb F_q)$; the only non-inner automorphism that we get this way is conjugation by $\begin{pmatrix} \epsilon & 0 \\ 0 & 1 \end{pmatrix}$, where $\epsilon$ is a non-square in $\mathbb F_q^\times$.

If an automorphism $\sigma$ of $\operatorname{SL}(2, \mathbb F_q)$ induces the identity on $\operatorname{PSL}(2, \mathbb F_q)$, then $g \mapsto g^{-1}\sigma(g)$ maps $\operatorname{SL}(2, \mathbb F_q) \to \{\pm1\}$. However, $\operatorname{SL}(2, \mathbb F_q)$ is generated by its unipotent elements, hence admits no non-trivial homomorphisms to a 2-torsion group (assuming $p \ne 2$). (EDIT: If $p = 2$, then $\operatorname{SL}(2, \mathbb F_q) = \operatorname{PSL}(2, \mathbb F_q)$, so there is no ambiguity.)

EDIT: I somehow misread the question as asking about the full automorphism group of $\operatorname{SL}(2, \mathbb F_q)$, not just the outer automorphism group; and it's always a good time to break out a result of Steinberg. As @RichardLyons and @YCor point out, since there are no diagram automorphisms in type $\mathsf A_1$, the outer automorphism group (of $\operatorname{SL}(2, \mathbb F_q)$, which, we argued above, is the same as that of $\operatorname{PSL}(2, \mathbb F_q)$) is $\langle\operatorname{Fr}\rangle \times \langle\operatorname{Int}\begin{pmatrix} \epsilon & 0 \\ 0 & 1 \end{pmatrix}\rangle \cong \operatorname C_n \times \operatorname C_2$, generated by the Frobenius and an appropriate conjugation in $\operatorname{PGL}(2, \mathbb F_q)$, when $p \ne 2$.

If $p = 2$, then, again as @YCor points out, there is no extra conjugation coming from $\operatorname{PGL}(2, \mathbb F_q)$ (since every element of $\mathbb F_q$ is a square), so we get that the outer automorphism group is just $\langle\operatorname{Fr}\rangle \cong \operatorname C_n$.

"Each automorphism $\sigma$ of $G$ can be written $\sigma = g f d i$, with $i$, $d$, $f$, and $g$ being inner, diagonal, field, and graph automorphisms, respectively" (Steinberg - Automorphisms of finite linear groups, 3.2). Here, as best as I can tell, $G$ is $\operatorname{PSL}(2, \mathbb F_q)$ (not $\operatorname{PGL}(2, \mathbb F_q)$); the definition of $G$ relies on a set that looks like $\mathfrak B$, whose definition I cannot find. A diagonal automorphism is one that arises by conjugation in the diagonal subgroup of $\operatorname{PGL}(2, \mathbb F_q)$, not just of $\operatorname{PSL}(2, \mathbb F_q)$; the only non-inner automorphism that we get this way is conjugation by $\begin{pmatrix} \epsilon & 0 \\ 0 & 1 \end{pmatrix}$, where $\epsilon$ is a non-square in $\mathbb F_q^\times$.

If an automorphism $\sigma$ of $\operatorname{SL}(2, \mathbb F_q)$ induces the identity on $\operatorname{PSL}(2, \mathbb F_q)$, then $g \mapsto g^{-1}\sigma(g)$ maps $\operatorname{SL}(2, \mathbb F_q) \to \{\pm1\}$. However, $\operatorname{SL}(2, \mathbb F_q)$ is generated by its unipotent elements, hence admits no non-trivial homomorphisms to a 2-torsion group (assuming $p \ne 2$).

"Each automorphism $\sigma$ of $G$ can be written $\sigma = g f d i$, with $i$, $d$, $f$, and $g$ being inner, diagonal, field, and graph automorphisms, respectively" (Steinberg - Automorphisms of finite linear groups, 3.2). Here, as best as I can tell, $G$ is $\operatorname{PSL}(2, \mathbb F_q)$ (not $\operatorname{PGL}(2, \mathbb F_q)$); the definition of $G$ relies on a set that looks like $\mathfrak B$, whose definition I cannot find. A diagonal automorphism is one that arises by conjugation in the diagonal subgroup of $\operatorname{PGL}(2, \mathbb F_q)$, not just of $\operatorname{PSL}(2, \mathbb F_q)$; the only non-inner automorphism that we get this way is conjugation by $\begin{pmatrix} \epsilon & 0 \\ 0 & 1 \end{pmatrix}$, where $\epsilon$ is a non-square in $\mathbb F_q^\times$.

If an automorphism $\sigma$ of $\operatorname{SL}(2, \mathbb F_q)$ induces the identity on $\operatorname{PSL}(2, \mathbb F_q)$, then $g \mapsto g^{-1}\sigma(g)$ maps $\operatorname{SL}(2, \mathbb F_q) \to \{\pm1\}$. However, $\operatorname{SL}(2, \mathbb F_q)$ is generated by its unipotent elements, hence admits no non-trivial homomorphisms to a 2-torsion group (assuming $p \ne 2$). (EDIT: If $p = 2$, then $\operatorname{SL}(2, \mathbb F_q) = \operatorname{PSL}(2, \mathbb F_q)$, so there is no ambiguity.)

EDIT: I somehow misread the question as asking about the full automorphism group of $\operatorname{SL}(2, \mathbb F_q)$, not just the outer automorphism group; and it's always a good time to break out a result of Steinberg. As @RichardLyons and @YCor point out, the outer automorphism group (of $\operatorname{SL}(2, \mathbb F_q)$, which, we argued above, is the same as that of $\operatorname{PSL}(2, \mathbb F_q)$) is $\langle\operatorname{Fr}\rangle \times \langle\operatorname{Int}\begin{pmatrix} \epsilon & 0 \\ 0 & 1 \end{pmatrix}\rangle \cong \operatorname C_n \times \operatorname C_2$, generated by the Frobenius and an appropriate conjugation in $\operatorname{PGL}(2, \mathbb F_q)$.

If $p = 2$, then, again as @YCor points out there is no extra conjugation coming from $\operatorname{PGL}(2, \mathbb F_q)$, so we just get the outer automorphism group $\langle\operatorname{Fr}\rangle \cong \operatorname C_n$.

"Each automorphism $\sigma$ of $G$ can be written $\sigma = g f d i$, with $i$, $d$, $f$, and $g$ being inner, diagonal, field, and graph automorphisms, respectively." (Steinberg, "Automorphisms of finite linear groups")

"Each automorphism $\sigma$ of $G$ can be written $\sigma = g f d i$, with $i$, $d$, $f$, and $g$ being inner, diagonal, field, and graph automorphisms, respectively" (Steinberg - Automorphisms of finite linear groups, 3.2). Here, as best as I can tell, $G$ is $\operatorname{PSL}(2, \mathbb F_q)$ (not $\operatorname{PGL}(2, \mathbb F_q)$); the definition of $G$ relies on a set that looks like $\mathfrak B$, whose definition I cannot find. A diagonal automorphism is one that arises by conjugation in the diagonal subgroup of $\operatorname{PGL}(2, \mathbb F_q)$, not just of $\operatorname{PSL}(2, \mathbb F_q)$; the only non-inner automorphism that we get this way is conjugation by $\begin{pmatrix} \epsilon & 0 \\ 0 & 1 \end{pmatrix}$, where $\epsilon$ is a non-square in $\mathbb F_q^\times$.

If an automorphism $\sigma$ of $\operatorname{SL}(2, \mathbb F_q)$ induces the identity on $\operatorname{PSL}(2, \mathbb F_q)$, then $g \mapsto g^{-1}\sigma(g)$ maps $\operatorname{SL}(2, \mathbb F_q) \to \{\pm1\}$. However, $\operatorname{SL}(2, \mathbb F_q)$ is generated by its unipotent elements, hence admits no non-trivial homomorphisms to a 2-torsion group (assuming $p \ne 2$).