L4.2

Let $$f(A) = (A^\text{T})^{-1}$$, because $$f(AB) = ((AB)^\text{T})^{-1} = (B^\text{T}A^\text{T})^{-1} = (A^\text{T})^{-1}(B^\text{T})^{-1} = f(A)f(B)$$ we see that $$f$$ is a homomorphism, and because $$f(A) = f(B) \iff (A^\text{T})^{-1} = (B^\text{T})^{-1}$$$$ \iff B^\text{T}(A^\text{T})^{-1}A^\text{T} = B^\text{T}(B^\text{T})^{-1}A^\text{T} \iff B^\text{T} = A^\text{T} \iff B = A$$ we furthermore see that $$f$$ is an automorphism.

It is however not an inner automorphism, i.e. it can not be written $$f(A) = BAB^{-1}$$ for any $$B \in \text{GL}_n(\mathbb{R}).$$ Let $$A = 2I$$ and let $$B$$ be an arbitrary invertible matrix, then $$f(A) = f(2I) = B2IB^{-1} = 2IBB^{-1} = 2I \not= ((2I)^\text{T})^{-1}.$$