Yes.  Observe first that $f$ can be first extended to an involution of $\mathbb{R}^3$ and then to an involution $F : S^3 \rightarrow S^3$ of the one-point compactification of $\mathbb{R}^3$.  A classical theorem of P.A. Smith then says that the fixed-point set of $F$ is homeomorphic to either $S^0$ or $S^1$ or $S^2$ or $S^3$; see Theorem 4 of

MR0000177 (1,30c)
Smith, P. A.
Transformations of finite period. II. 
Ann. of Math. (2) 40, (1939). 690–711. 

By the way, the proof shows that if $F$ is orientation-preserving, then the fixed-point set of $F$ must either be $S^1$ or $S^3$, while if $F$ is orientation-reversing, then the fixed-point set of $F$ must either be $S^0$ or $S^2$.  In any case, from our construction it is clear that the fixed-point set of $F$ must be $S^3$, i.e. $F = \text{id}$.

I should remark that Smith's theorem is the beginning of a long story.  See, in particular, the book

MR0758459 (86i:57002)
The Smith conjecture. 
Papers presented at the symposium held at Columbia University, New York, 1979. Edited by John W. Morgan and Hyman Bass. Pure and Applied Mathematics, 112. Academic Press, Inc., Orlando, FL, 1984. xv+243 pp. ISBN: 0-12-506980-4 

The main theorem discussed in this book says that if $F$ is a nontrivial periodic orientation-preserving homeomorphism of $S^3$, then the fixed-point set of $F$ is an *unknotted* circle; this implies that $F$ is topologically conjugate to an element of the orthogonal group.  This result was one of the first triumphs of Thurston's work on 3-manifolds.