It is fairly easy to see from the formalism of the L group that the L group of a quasisplit unitary group will be a nontrivial semidirect product of $GL\_n(\mathbb{C})$ (for the appropriate value of $n$), and a group of order two. When $n$ is odd this determines the isomorphism class uniquely. When $n$ is even, it does not. Indeed, an outer automorphism is of the form $g \mapsto A{}^tg^{-1}A^{-1}$ for some $A \in GL\_n(\mathbb C).$ It is of order two if and only if $A$ is either symmetric or skew-symmetric. It seems to me that in the literature the $L$ group of a quasisplit unitary group is consistently defined using a skew-symmetric $A.$ Can anyone explain why this is correct?



Why is the L group of an even unitary group what it is?
There are three separate issues here: exactly which unitary or special unitary groups you're thinking about (in effect: exactly which non-degenerate hermitian spaces), how to define $L$-groups in general, and how that definition works out in the quasi-split case.


In general, if $G$ is a pinned split connected reductive group over a field $k$ and it has based root datum $(R, \Delta)$ arising from the pinning then via the notion of pinned automorphism we have
$${\rm{Aut}}(G\_{k\_s}) = G^{\rm{ad}}(k\_s) \ltimes {\rm{Aut}}(R, \Delta)$$
compatible with the action of ${\rm{Gal}}(k\_s/k)$.
In this way, the pointed set ${\rm{H}}^1(k, {\rm{Aut}}(R,\Delta))$ of conjugacy classes of actions of ${\rm{Gal}}(k\_s/k)$ on $(R, \Delta)$ is in natural bijection with the set of $k$-isomorphism classes of quasi-split $k$-forms of $G$, and every $k$-form of $G$ has a unique quasi-split inner form. 


For example, if $G = {\rm{SL}}\_n$ for $n>1$ then it is simply connected with diagram A$\_{n-1}$, so the split form is the only quasi-split form when $n=2$ (as A$\_1$ has no nontrivial automorphisms) whereas otherwise the diagram automorphism group has order 2 and so the set of non-split quasi-split $k$-forms of ${\rm{SL}}\_n$ is classified up to isomorphism by the set of quadratic Galois extensions $k'/k$ up to isomorphism. The link is that the group attached to $k'/k$ becomes split over $k'$. But what is it explicitly? That is where you preferred (special) unitary groups come in. 


So now assume $n\ge 3$ and consider the hermitian space $({k'}^n, h\_n)$ for a quadratic etale $k$-algebra $k'$ (i.e., a quadratic Galois field extension, or $k \times k$) with unique non-trivial automorphism denoted $z \mapsto \overline{z}$ and the non-degenerate hermitian form $h\_n$ defined by 
$$h\_{2m}(\vec{x}, \vec{y}) = \sum\_{j=1}^{m}(x\_j \overline{y}\_{m+j}+x\_{m+j}\overline{y}\_j)$$
for $n = 2m$ and
$$h\_{2m+1} = x\_0 \overline{y}\_0 + h\_{2m}((x\_1,\dots,x\_{2m}),(y\_1,\dots,y\_{2m}))$$
for $n=2m+1$. There are very many other kinds of rank-$n$ non-degenerate hermitian spaces over $k$ in general. 


If $k' = k \times k$ then ${\rm{SU}}(k'/k, h\_n)$ turns out to be ${\rm{SL}}\_n$ by another name; see Exercise 3 at <http://math.stanford.edu/~conrad/252Page/homework/hmwk7.pdf>
(where ${\rm{SU}}(h)$ for hermitian spaces $(V',h)$ over quadratic etale algebras $k'/k$ are described inside the Weil restriction ${\rm{R}}\_{k'/k}({\rm{SL}}(V'))$). I claim that for $k'/k$ a quadratic Galois field extension, the $k$-group ${\rm{SU}}(k'/k, h\_n)$ is the unique non-split quasi-split form of ${\rm{SL}}\_n$ that splits over $k'$. 


If you look at Example 7.1.10 (and Example 7.1.5) in the article "Reductive Groups Schemes" from the 2011 Luminy summer school on SGA3 (and set the base $S$ there to be ${\rm{Spec}}(k)$) you'll see a description for any $n>2$ of the Galois-twisting encoding exactly the unique non-split quasi-split $k$-form $G$ of ${\rm{SL}}\_n$ that is split by $k'/k$, using the pinning associated to the upper-triangular Borel subgroup $B$ and the order-2 automorphism of ${\rm{SL}}\_n$ given by $g \mapsto w({}^tg^{-1})w^{-1}$ where $w$ is the anti-diagonal matrix with alternating $1$'s and $-1$'s beginning with $1$ in the upper-right (and the role of transpose is to ensure that $B\_{k'}$ is stable under the Galois descent, so the associated $k$-form really is quasi-split). From that explicit description of $G$ one describes $G(A)$ inside ${\rm{SL}}\_n(k' \otimes\_k A)$ for any $k$-algebra $A$ as fixed-points of a specific involution, and you can work out that this coincides *exactly* with the description of ${\rm{SU}}(k'/k, h\_n) \subset {\rm{R}}\_{k'/k}({\rm{SL}}\_n)$ (on $A$-points, for any $A$).


So we have learned two things, for all $n > 2$ on equal footing: the groups ${\rm{SU}}(k'/k, h\_n)$ for varying quadratic Galois $k'/k$ really are characterized in terms of the quasi-split property over $k$ and the split property over $k'/k$ as claimed, and we see exactly how they are made from ${\rm{SL}}\_n$ via an explicit involutive automorphism over $k'$. We likewise conclude that ${\rm{U}}(k'/k, h\_n)$ admits the same Galois-twisting description in terms of ${\rm{GL}}\_n$ and $k'/k$ for each $n > 2$. This Galois-twisting description with an explicit involution is what is going to answer your actual question concerning the $L$-group for ${\rm{U}}(k'/k, h\_n)$. But first we have to recall what exact is the definition of the $L$-group.


Strictly speaking, the uniform definition of the $L$-group treating all connected reductive groups $G$ over a local or global field $k$ on an equal footing involves a semi-direct product against the *entire* Weil group $W\_k$ of $k$ for a specific action of $W\_k$ on the connected reductive $\mathbf{C}$-group $^{L}G^0$ with the dual based root datum, so it has infinitely many connected components. But in practice one passes to the quotient by the finite-index open kernel of the $W\_k$-action to collapse it down to a semi-direct product against a finite Galois group ${\rm{Gal}}(k'/k)$. It is the latter viewpoint you're thinking of, but really that is not the actual definition of the $L$-group (though in practice it works just as well). However you slice it, the real issue is: what is the $W\_k$-action on ${}^LG^0(\mathbf{C})$? Or in our case of interest with $G = {\rm{U}}(k'/k, h\_n)$ with $n \ge 3$ and $k'/k$ a quadratic Galois field extension, where it collapses down to an action of ${\rm{Gal}}(k'/k)$ on ${\rm{GL}}\_n(\mathbf{C})$ (really the "dual"), what is the associated involution of ${\rm{GL}}\_n(\mathbf{C})$? I claim that it is *exactly* the same recipe which we saw in the very description of $G$ as a quasi-split form of ${\rm{GL}}\_n$ obtained by $k'/k$-twisting against the involution $g \mapsto w({}^tg^{-1})w^{-1}$ of ${\rm{GL}}\_n$ over $k'$.


In general, the classification of connected reductive groups over fields involves Tits' notion of the $\ast$-action of the absolute Galois group on the based root datum, a notion that simplifies a lot in the quasi-split case (in that a certain intervention of the Weyl group becomes trivial): see section 12.2 of the course notes <http://math.stanford.edu/~conrad/249BW16Page/handouts/249B_2016.pdf> for the $\ast$-action and how it works out in the quasi-split case in terms of the explicit Galois-twisting that makes the group from the split form. Coming back to ${\rm{U}}(k'/k, h\_n)$, for which we worked out the $k'/k$-twisting to get it from ${\rm{GL}}\_n$ above, this gives exactly the recipe you wanted **provided** that we can sort out the role of the identification of ${\rm{GL}}\_n$ as its own dual (to rigorously justify the correctness of how we describe the action on the $\mathbf{C}$-group ${}^L{\rm{GL}}\_n$). Passing to the dual torus lattice and its dual basis of coroots (due to how ${}^LG^0$ is *defined* as a pinned $\mathbf{C}$-group), we get desired formula on the associated *pinned* "dual ${\rm{GL}}\_n$" because the automorphism $f:g \mapsto w({}^tg^{-1})w^{-1}$ satisfies
$${}^t (f({}^tg)) = {}^t(w({}^t({}^tg)^{-1})w^{-1}) = ({}^tw^{-1})({}^tg^{-1})({}^tw) = w({}^tg^{-1})w^{-1} = f(g);$$
the second-to-last equality uses that ${}^tw = (-1)^{n-1}w=w^{-1}$.


