$f(x)= {}^tx^{-1}$ is an automorphism of GL$_n(\mathbb{R})$ This is a question from the free Harvard online abstract algebra lectures.  I'm posting my solutions here to get some feedback on them.  For a fuller explanation, see this post.
This problem is assigned in the 4th lecture.

Define $f:GL_n(\mathbb{R}) → GL_n(\mathbb{R})$ by $f(A)=^tA^{-1}$ (where $^tA$ is the transpose of $A$).  Show that $f$ is an automorphism, but not an inner automorphism for n ≥ 1.

Let $A,B\in GL_n(\mathbb{R})$.  Then
$f(AB)={}^t(AB)^{-1}=({}^tB^tA)^{-1}={}^tA^{-1}\cdot{}^tB^{-1}=f(A)\cdot f(B)$.  So $f$ is a homomorphism.  For any $A \in GL_n(\mathbb{R}), f(f(A))=f({}^tA^{-1})={}^t({}^tA^{-1})^{-1}=A$.  So $f^{-1}=f$.  Since $f$ has an inverse for all $A\in GL_n(\mathbb{R})$, $f$ is bijective.  Therefore, $f$ is an automorphism of $GL_n(\mathbb{R})$.
Assume $f$ is an inner automorphism.  Let $A=\lambda I_n$.  Since $A$ is in the center of $GL_n(\mathbb{R})$, there is some $B \in GL_n(\mathbb{R})$ such that $f(A)=BAB^{-1} = BB^{-1}A=A$.  But ${}^tA^{-1}=\frac{1}{\lambda}I_n\neq A$ for $\lambda\neq1$.  Therefore, $f$ is not an inner automorphism.
Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem.
Thanks.
 A: The math all looks fine.  Here are some style critiques (bear in mind that style critiques are somewhat subjective.)


*

*Although the order does not matter, I bristle at seeing the $t$ and $-1$ on either side of a matrix without an indication of which is done first.  In the lectures, presumably this exercise appears after some discussion where it is proved that the order is immaterial, but divorced from such a context, it looks a little weird.  (If only because the proof that the order is immaterial is so similar to the proof that $f$ is multiplicative--- both are algebraic calculations involving a delicate balance of transposing and inversing, and how these operations individually treat products.)

*I would write "So $f$ is invertible, and in fact $f^{-1} = f$" rather than "So $f^{-1} = f$."  (General principle: if $f$ has not yet been explicitly announced as invertible, do not write $f^{-1}$.)  If you don't buy that general principle, I would still say: at this stage in the argument, the very existence of $f^{-1}$ has just been established, and this deserves more notice than the fact that $f^{-1} = f$, because it is the existence of the inverse that really establishes that $f$ is an automorphism, and not so much the specific fact that $f^{-1}$ happens to be $f$.

*Depending on how "automorphism" has been defined--- it may have been defined to be a homomorphism that has an inverse in the sense that you just established $f$ did--- it may not be necessary to detour through bijectivity to get to the fact that $f$ is an automorphism.  In any case, "Since $f$ has an inverse for all $A \in \operatorname{GL}_n(\mathbb{R})$" should at least be replaced with just "Since $f$ has an inverse", because the "for all $A \in \operatorname{GL}_n(\mathbb{R})$" is meaningless here.  (The map $f$ either has an inverse or does not; "has an inverse" is not a parametrized statement that depends on a matrix $A$.)

*I would re-order the discussion in the last paragraph a little bit.  The issue for me is the distance between restatements of assumptions and where they are used.  "There is some $B \in \operatorname{GL}_n(\mathbb{R})$ with $f(A) = BAB^{-1}$" is where the assumption that $f$ is an inner automorphism is being used, but as currently written, this assertion is introduced by the statement "Since $A$ is in the center of $\operatorname{GL}_n(\mathbb{R})$".  This does not explain why there is that $B$, but rather why the calculational step following that assertion (namely the equality $BAB^{-1} = BB^{-1}A$) is valid.  

*In "Let $A = \lambda I_n$" you are implicitly introducing the number $\lambda$, but not telling the reader anything else about it.  On finishing this sentence, a reader may be unsure whether (a) you are intending to choose a specific value of $\lambda$ later, or (b) you intend to do something for an arbitrary value of $\lambda$, or (c) you have already fixed a value of $\lambda$ somewhere, and the reader accidentally skipped over it.  
The item (c) is the most important one to be mindful of in situations like this--- if not when writing individual exercise solutions, in longer-form mathematical writing.  Wherever possible, when introducing a new symbol, make clear that you are introducing it, so the reader doesn't re-scan what they've just read looking for something they missed.
You could, for example, say "Fix any nonzero number $\lambda$ and consider $A = \lambda I_n$.  From the definition of $f$ we have $f(A) = \cdots = \lambda^{-1} I_n$, and in particular if we choose $\lambda \neq \pm 1$ we see that $f(A) \neq A$.  On the other hand, if $f$ were inner we would have [...]  So $f$ is not inner."  [Note that $\lambda^{-1} = \lambda$ holds for $\lambda = -1$ also; I guess this is a tiny math error in the original.]
Alternatively, you could avoid the symbol $\lambda$ entirely and just do the calculation for one particular value of $\lambda$, e.g. $2$.  This is where people fall into two camps.  Some would say that it is bad to randomly introduce $2 I_n$ to the discussion without any indication of where the choice "came from", and that the discussion with general $\lambda$ followed by a specialization to $\lambda \neq \pm 1$ leaves the reader with more understanding.  Others would say that as the goal is only to show by example that $f$ is not inner, then it complicates things to introduce a parameter into the discussion when one can give just a single example.
A: This was my attempt to show $f$ does not belong to the inner automorphism group of $GL_n(\mathbb R)$.
If $f \in \textrm{Inn}(GL_n(\mathbb R))$, it could be written:
\begin{align} f(A)&=gAg^{-1}\\ \quad {}^tA^{-1} &= gAg^{-1} \quad (1)
\end{align} for some $g \in GL_n(\mathbb R)$.
To show this is not possible, let us take $\textrm{det}(\cdot)$ on both sides of $(1)$:
\begin{align}  \textrm{det}({}^tA^{-1})&=\textrm{det}(gAg^{-1})\\
\textrm{det}(A^{-1})&=\textrm{det}(g) \cdot \textrm{det}(A) \cdot \textrm{det}(g^{-1})\\
\textrm{det}(A)^{-1} &= \textrm{det}(g) \cdot \textrm{det}(A) \cdot \textrm{det}(g)^{-1}\\
\textrm{det}(A)^{-1} &= \textrm{det}(A) \leadsto \textrm{det}(A) = \pm 1
\end{align}
So equation $(1)$ does not hold unless it is for an $A$ such that $\textrm{det}(A) = \pm 1$, which is only a subset of $GL_n(\mathbb R)$
$(1)$ may only be true for $A\in S:= \{ M|\textrm{det}(M)=\pm1 \}\subset GL_n(\mathbb R)$, and therefore $f \notin \textrm{Inn}(GL_n(\mathbb R))$.
$A\in S$ is a sufficient but not necessary condition for $(1)$ to be true; interesting to see that with the $A=\lambda I$ you take, $(1)$ is true when $\lambda = \pm 1$, which is also when $\textrm{det}(A) = \pm 1$.
