I am confused with morphisms of supermanifolds. Take a simple example $f:R^{01}\to R^{01}$. By (one of) definition, $f$ is a morphism of superalgebras of functions $C(R^{01})\to C(R^{01})$. Morphisms of superalgebras preserve the grading, I deduced that $f$ have the form $1\mapsto 1, \theta\mapsto x\theta$, i.e. $Hom(R^{01}\to R^{01})=R^1$ (as a set?). But I read from a paper that $Hom(R^{01},R^{01})=R^{11}$. What is going on? Thanks in advance!
You are right that the set of supermanifold morphisms $Hom(\mathbb R^{01},\mathbb R^{01})$ to itself is $\mathbb R^1$. However, one can define for supermanifolds $X,Y$ with $\dim X=0d$ a supermanifold $map(X,Y)$ of morphisms from $X$ to $Y$, by $Hom(Z,map(X,Y))=Hom(Z\times X,Y)$ for all supermanifolds $Z$. And $map(\mathbb R^{01},\mathbb R^{01})=\mathbb R^{11}$. 


@Ma: As an answer to your following question:
Take a look at arXiv:math/0307303, where this question is discussed. For $d=1$, it is wellknown (and due to Kontsevich, I think), that $map(R^{01},M)$ is the total space of the odd tangent bundle $\Pi TM$ of $M$. If $U$ is a superdomain of dimension $pq$, then $map(R^{0d},U)=U\times R^{pr+qsps+qr}$ where $(r+1)s=2^{d1}2^{d1}$ is the graded dimension of $\bigwedge R^d$. This you can check using the definition of $map$ and the characterisation of morphisms of supermanifolds as given in Leites. Another good source on this subject (for $d=1$), is the paper "Differential forms and 0dimensional supersymmetric field theories" by Hohnhold, Kreck, Stolz and Teichner. 

