Let $E/F$ be a finite field extension. Let $G\_E$ be a reductive linear algebraic group defined over $E$ and let $G=\mathrm{R}\_{E/F}G\_E$ be the Weil restriction of scalars. Then $G$ is a linear algebraic group defined over $F$ and $\mathrm{R}\_{E/F}$ is a functor from the category of linear algebraic groups defined over $E$ to the category of linear algebraic groups defined over $F$. 


Consider $\mathrm{Aut}\_E(G\_E)$, the group of automorphisms of $G\_E$ as an algebraic group over $E$. The group $\mathrm{Aut}\_F(G)$ is defined similarly. The functor provides a homomorphism 
$$
\mathrm{R}\_{E/F}:\mathrm{Aut}\_E(G\_E)\to\mathrm{Aut}\_F(G).
$$
My question is, what is the quotient
$$
\Gamma=\mathrm{Aut}\_F(G)/\mathrm{R}\_{E/F}(\mathrm{Aut}\_E(G\_E))?
$$


If $G\_E=\mathrm{GL}(n)$, then I would like to know in particular that $\Gamma$ is the Galois group $\mathrm{Gal}(E/F)$. To see concretely how the Galois group acts, we can construct the restriction of scalars by choosing an $F$-basis for $E$ that provides a map from $G\_E(E)$ to $\mathrm{GL}\_N(F)$ where $N=n[E:F]$. The image of this map defines an algebraic group over $F$ which represents the restriction of scalars and the Galois group action on $G\_E(E)$ can be transferred to an action on the restriction of scalars. 



Automorphism of restriction of scalars
You surely meant to assume the finite extension of fields $E/F$ is separable (otherwise ${\rm{R}}\_{E/F}(H)$ is *never* reductive for a smooth connected affine $E$-group $H \ne 1$). Also, this is one of those cases where more generality clarifies the situation. The reason I say this is that by limiting yourself only to Weil restriction from a *field* over $F$ rather than more generally from a finite etale $F$-algebra (i.e., a product of several fields over $F$, all finite and separable over $F$), you prevent yourself from accessing the technique of scalar extension on $F$ to simplify computations.


So let's pose the situation more broadly as follows. Let $F$ be a field, $E$ and $E'$ nonzero finite etale $F$-algebras, and $G$ and $G'$ smooth affine groups over $E$ and $E'$ respectively such that their fibers over the factor fields of $E$ and $E'$ are connected reductive and *non-trivial*. We want to describe $F$-isomorphisms $f: {\rm{R}}\_{E/F}(G) \simeq {\rm{R}}\_{E'/F}(G')$ in terms of more concrete data, perhaps under some hypotheses on the groups (and in particular to address the case when $E = E'$ is a field and $G = G' \ne 1$). 


Given 
an $F$-algebra isomorphism $\alpha:E \simeq E'$ and a group isomorphism $\varphi:G \simeq G'$ over $\alpha$ there is an evident associated $F$-group isomorphism ${\rm{R}}\_{\alpha/F}(\varphi): {\rm{R}}\_{E/F}(G) \simeq
{\rm{R}}\_{E'/F}(G')$. So we can ask the (easy) question of whether the pair $(\alpha, \varphi)$ is uniquely determined by 
the $F$-isomorphism ${\rm{R}}\_{\alpha/F}(\varphi)$, and the more serious question of whether every $f$ arises from such a pair. The first key point, and the reason for recasting the setup in terms of the generality of (nonzero) finite etale $F$-algebras rather than only *fields* (of finite degree and separable) over $F$ is that if $F'/F$ is any extension field (such as a separable closure, or simply a big enough finite Galois extension, or whatever) then
$${\rm{R}}\_{E/F}(X) \otimes\_F F' \simeq {\rm{R}}\_{(E \otimes\_F F')/F'}(X \otimes\_E (E \otimes\_F F'))$$
for any affine $E$-scheme $X$ of finite type. Note that if $E$ were a field then $E \otimes\_F F'$ is typically *not* a field, but that since $E$ is finite etale over $F$ at least $E \otimes\_F F'$ is finite etale over $F'$. More specifically, a nonzero finite etale $F$-algebra $E$ always has the form $E = \prod E\_i$ for fields $E\_i$ that are finite separable over $F$, any scheme $X$ over ${\rm{Spec}}(E) = \coprod {\rm{Spec}}(E\_i)$ always has the form $X = \coprod X\_i$ where $X\_i$ is the $E\_i$-fiber of $X$, and for $X$ affine and finite type over $E$ we have
$${\rm{R}}\_{E/F}(X) \simeq \prod {\rm{R}}\_{E\_i/F}(X\_i);$$
this makes contact back with the more familiar world of Weil restriction through field extensions, but the language of finite etale $F$-algebras is much more efficient for bookkeeping purposes with the notation when we want to bring in ground field extensions (as we will do below!).


Let's now come back to the question of whether an $F$-isomorphism
$f: {\rm{R}}\_{E/F}(G) \simeq {\rm{R}}\_{E'/F}(G')$ having the form ${\rm{R}}\_{\alpha/F}(\varphi)$ uniquely determines the $F$-algebra isomorphism $\alpha$ and the group isomorphism $\varphi$ over $\alpha$.
Since we're only asking about uniqueness of such a pair, not existence, it suffices to check after making a ground field extension! And we were careful to set up our entire framework in a manner that is not destroyed by such an operation, so for the purpose of proving uniqueness we can apply ground field extension from $F$ to $F\_s$ so that $F$ is separably closed and hence all finite etale $F$-algebras are split!


More to the point, now (for the purpose of proving uniqueness) we can assume as $F$-algebras that $E = F^I$ and $E' = F^J$ for non-empty finite sets $I$ and $J$, so $\alpha$ is nothing other than a bijection of sets $\tau:I \simeq J$ (with $\alpha((x\_i)) = (x\_{\tau^{-1}(j)})$). Likewise, now $G = \coprod G\_i$ and $G' = \coprod G'\_j$ with $G\_i$ and $G'\_j$
all *non-trivial*, and
$\varphi$ is nothing other than a collection of $F$-isomorphisms
$\varphi\_i:G\_i \simeq G'\_{\tau(i)}$ for $i \in I$. Since ${\rm{R}}\_{E/F}(G) = \prod G\_i$ and ${\rm{R}}\_{E'/F}(G') = \prod G'\_j$,
the uniqueness question comes down to the concrete problem of whether the product isomorphism of $F$-groups 
$$\prod\_{i \in I} G\_i \simeq \prod\_{j \in J} G'\_j$$
defined by $(g\_i) \mapsto (\varphi\_{\tau^{-1}(j)}(g\_{\tau^{-1}(j)}))$ uniquely determines both $\tau$ and the collection of $\varphi\_i$'s.
Since all $G\_i$ and $G'\_j$ are nontrivial, the uniqueness of both is immediate by chasing the image of $G\_i$ under the product isomorphism.


That settles the uniqueness aspect, so now we can turn to the much more interesting existence aspect. Obviously this cannot be affirmative in general (just imagine that $G$ is itself a Weil restriction, and consider that a composition of Weil restrictions is a Weil restriction for the composite extension). So the good result is as follows:


**Theorem**. *Assume that the connected reductive fibers of $G$ and $G'$ over the respective factor fields of $E$ and $E'$ are semisimple, absolutely simple, and simply connected. Then any $F$-group isomorphism
${\rm{R}}\_{E/F}(G) \simeq {\rm{R}}\_{E'/F}(G')$ has the form ${\rm{R}}\_{\alpha/F}(\varphi)$ for a $($uniquely determined$)$ pair
$(\alpha, \varphi)$ as above. The same holds with "adjoint type" in place of "simply connected".*


The "absolutely simple" hypothesis rules out the obstruction caused by internal Weil restrictions as noted above, the "semisimple" condition is needed prior to making any meaningful/useful notion of "absolutely simple", and the "simply connected" (or alternatively "adjoint type") hypothesis ensures that the absolute root datum (not just root system!) is a direct product of those for its irreducible components. 


For a proof of the above Theorem (including a reference to its historical antecedent in the older language of Weil restriction only through field extensions, at the cost of slightly heavier notation), see Proposition A.5.14 in the book *Pseudo-reductive Groups*. 


Now let's finally come back to the original question concerning automorphisms of $H := {\rm{R}}\_{E/F}({\rm{GL}}\_n)$ for a finite separable extension of fields $E/F$ and $n \ge 1$. For $n = 1$ it is really hopeless to say anything in general (since passing to the Cartier dual Galois lattice ${\rm{Ind}}\_{\Gamma\_F}^{\Gamma\_E}(\mathbf{Z})$ shows that (if $E \ne F$) there are a *lot* of automorphisms, as we can see by first considering the situation after scalar extension to $\mathbf{Q}$). So let's assume $n \ge 2$. Then the derived group $\mathscr{D}(H)$ coincides with ${\rm{R}}\_{E/F}({\rm{SL}}\_n)$, and we have
$$H \simeq (\mathscr{D}(H) \times Z\_H)/\mu = ({\rm{R}}\_{E/F}({\rm{SL}}\_n) \times {\rm{R}}\_{E/F}({\rm{GL}}\_1))/{\rm{R}}\_{E/F}(\mu\_n)$$
where $Z\_H$ is the (scheme-theoretic) center of $H$ and $\mu = \mathscr{D}(H) \cap Z\_H
= Z\_{\mathscr{D}(H)} \simeq {\rm{R}}\_{E/F}(\mu\_n)$.


Thus, any $F$-automorphism $f$ of $H$ is exactly the data of an $F$-automorphism $f\_1$ of $\mathscr{D}(H) = {\rm{R}}\_{E/F}({\rm{SL}}\_n)$ and an $F$-automorphism $f\_2$ of $Z\_H = {\rm{R}}\_{E/F}({\rm{GL}}\_1)$ that coincide on their overlap ${\rm{R}}\_{E/F}(\mu\_n)$. The preceding stuff shows that $f\_1$ is uniquely of the form ${\rm{R}}\_{\alpha/F}(\varphi)$ for an $F$-algebra automorphism $\alpha$ of $E$ and a group automorphism $\varphi$ of ${\rm{SL}}\_n$ over $\alpha$. Our knowledge of the automorphism group of ${\rm{SL}}\_n$ over *fields* shows that the effect of $\varphi$ on $\mu\_n$ is (up to the effect of $\alpha$) either trivial or inversion, and *both* options extend to automorphisms of ${\rm{GL}}\_1$ over $\alpha$! Thus, whatever $f\_1$ we might want to consider, a compatible $f\_2$ can *always* be found. 


Consequently, we see that the obstruction to an affirmative answer to your question is *exactly* the group of $F$-automorphisms of the induced torus ${\rm{R}}\_{E/F}({\rm{GL}}\_1)$ restricting to the identity on its $n$-torsion subgroup ${\rm{R}}\_{E/F}(\mu\_n)$. Since $n > 1$, whenever $E \ne F$ we can make infinitely many such extra automorphisms. I conjecture that you posed the question for ${\rm{GL}}\_n$ under the mistaken belief that it would make the problem easier (whereas now you see that the central torus actually makes the situation harder), and that the positive answer for ${\rm{SL}}\_n$ (and more generally as in the Theorem above) is the sort of thing with which you would be satisfied. So I'll leave it to you to mull over the extra junk automorphisms arising from the central torus if you really do need that information (which feels doubtful).


