Notion of uniquely transportable Categories I do not fully understand the notion of uniquely transportable categories. In the “Abstract and Concrete Categories” book one says that, for example, concrete category $(\mathbf G\mathbf r\mathbf p,U)$ over $\mathbf X$ with forgetful functor is uniquely transportable.
But take, for example, the group of order 4 with such composition law:
$a^2=b$
$ab=c=a^3$
$a^4=b^2=e$.
And take the $\mathbf X-$isomorphism which maps $e\rightarrow a, a\rightarrow b, b\rightarrow c, c\rightarrow e$, then we have to find a group $B$ of order 4, which is isomorphic to one mentioned above.
And the question is: how can we do this for case mentioned above and for arbitrary group $D\in \mathbf G\mathbf r\mathbf p$?

Given any group $D$ and a set $E$ with a bijection $f:D\to E$, make $E$ a group using the operation $*$ defined by $x*y=f(f^{-1}(x)\cdot f^{-1}(y))$, where $\cdot$ is the multiplication of $D$.  The fact that $(E,*)$ is a group follows easily from the fact that $(D,\cdot)$ is a group and $f$ is a bijection.  And essentially by definition of $*$, $f$ is an isomorphism from $(D,\cdot)$ to $(E,*)$.
In your case, you just apply this to your bijection.  Explicitly, $a$ is the identity element of $*$, $b*b=c$, $b*c=e=b*b*b$, and $b*b*b*b=c*c=a$.
