The phrase "degrees of freedom" refers to the number of independent real number valued parameters (as does the more formal term "dimension").
So, for example, each orientation preserving isometry of $n$-dimensional Euclidean space $E^n$ is a composition $T \circ O$ of a unique orthogonal transformation $O$ followed by a unique translation $T$, and these two factors are independent of each other. Translations are parameterized by vectors which have $n$ degrees of freedom, and rotations of $E^n$ are parameterized by orthogonal $n \times n$ matrices which have $\frac{(n-1)n}{2}$ degrees of freedom. Since the degrees of freedom of translations and of rotations are independent of each other, together they have $n + \frac{(n-1)n}{2}=\frac{n(n+1)}{2}$ degrees of freedom.
Now let's bring in reflections. Let me use $R_n$ to denote the reflection in the coordinate plane $x_n=0$. A general orientation reversing isometry can be expressed uniquely in the form $T \circ O \circ R_n$ where $T,O$ are independently chosen translation and orthogonal transformation. It follows that a general isometry can be written uniquely in the form $T \circ O \circ (R_n)^e$ where we independently choose three things: the translation $T$ with $n$ degrees of freedom; the orthogonal rotation $O$ with $\frac{(n-1)n}{2}$ degrees of freedom; and the exponent $e \in \{0,1\}$.
The choice of the exponent $e$ is discrete --- either $0$ or $1$ --- and this does not represent a "degree of freedom" because it is not a real number valued parameter.
By the way, what makes the calculations in my answer work correctly is the fact that the decomposition $T \circ O \circ (R_n)^e$ is uniquely determined by the isometry. In your question, where you use the fact that every isometry can be written as a composition of reflections, that composition is not unique, and so using it to count degrees of freedom can lead to errors.
Just as an example, picking any even number $n \ge 2$, every translation of the line can be written as a product of $n$ reflections. Since a reflection of the line has $1$ degree of freedom (namely, the reflection point), this would seem to be a proof that the translations of the line have dimension $n$. Since $n$ is arbitrary, there is an error in this argument... which I will leave you to ponder.
According to Schwartzman's The Words of Mathematics:
symmetric (adjective), symmetry (noun): the first element is from Greek sun- "together with," from the Indo-European root ksun "with." The second element is from Greek metron "a measure." The Indo-European root is probably me- "to measure." Suppose two points are symmetric with respect to a line; if you measure the distance between one of the points and the line of symmetry, then "together with" that measurement you have simultaneously also measured the distance between the other point and the line of symmetry; the two distances are equal.
So the root of the word concerns how multiple things share the same measurement. This concept is itself symmetric under exchanging the things being measured. There is no preference for transforming $x\mapsto y$ over the reverse direction $y\mapsto x$; both should be equally possible.
Best Answer
Who says it isn't? In the quoted wiki article we read:
Most generally the symmetry group of an object is just a set of all auto isomorphisms of that object. The word "relevant" is crucial here and in different context it means different things.
If you look at the square as a set then isomorphism = bijection and your example is valid. If you look at it as a topological space then isomorphism = homeomorphism and your example fails cause switching 2 points is not continuous. If you look at it as a subset of some vector space then isomorphism may mean linear isomorphism. If it is metric space then it may mean isometry and so on and so on...