Answer
See below
Work Step by Step
a) We know that $1.A=A.1 $
and $1=1^{-1} \\
\rightarrow 1.A=A.1\\
A=1^{-1}.A.1$
Hence, a matrix $A$ always similar to itself.
b) We have $A$ is similar to $B$
then $B=S^{-1}AS\\
\rightarrow A=SBS^{-1} \\
\rightarrow A=(S^{-1})^{-1}BS^{-1}$
Since $S$ is invertible, $S^{-1}$ is also invertible.
Hence, B is similar to A.
c) Since $A$ is similar to $B$, we have $B=S^{-1}AS$
and $B$ is similar to $C$, $\rightarrow C=P^{-1}BP\\
\rightarrow C=P^{-1}(S^{-1}AS)P\\
\rightarrow C=(SP)^{-1}ASP $
Hence, $A$ is similar to $C$
