There is a well known result that if $P$ is an abelian $p$-group of rank $2$ then $$|Aut(P)|=(p-1)^kp^j(p+1)^r,$$ for some $k,j,r$ which depend on the order of $P$.

Moreover, if $P=C_{p^t}\times C_{p^s}$ for $t\neq s$ then $r=0$ in the above equation.

However, I didn't find any reference to these facts.

There is a result saying that if $P$ is an abelian $p$-group of rank $2$ then $$|Aut(P)|=(p-1)^kp^j(p+1)^r,$$ for some $k,j,r$ which depend on the order of $P$.

Moreover, if $P=C_{p^t}\times C_{p^s}$ for $t\neq s$ then $r=0$ in the above equation.

However, I didn't find any reference to these facts.

There is a well known result that if $P$ is an abelian $p$-group of rank $2$ then $$|Aut(P)|=(p-1)^kp^j(p+1)^r,$$ for some $k,j,r$ which depend on the order of $P$.

Moreover, if $P=C_{p^t}\times C_{p^s}$ for $t\neq s$ then $j=0$ in the above equation.

However, I didn't find any reference to these facts.

There is a well known result that if $P$ is an abelian $p$-group of rank $2$ then $$|Aut(P)|=(p-1)^kp^j(p+1)^r,$$ for some $k,j,r$ which depend on the order of $P$.

Moreover, if $P=C_{p^t}\times C_{p^s}$ for $t\neq s$ then $r=0$ in the above equation.

However, I didn't find any reference to these facts.

There is a well known result that if $P$ is an abelian $p$ group of of rank $2$ then $$|Aut(P)=(p-1)^kp^j(p+1)^r,$$ for some $k,j,r$ which depend on the order of $P$.

Moreover, if $P=C_{p^t}\times C_{p^s}$ for $t\neq s$ then $j=0$ in the above equation.

However, I didn't found any reference to these facts.

There is a well known result that if $P$ is an abelian $p$-group of rank $2$ then $$|Aut(P)|=(p-1)^kp^j(p+1)^r,$$ for some $k,j,r$ which depend on the order of $P$.

Moreover, if $P=C_{p^t}\times C_{p^s}$ for $t\neq s$ then $j=0$ in the above equation.

However, I didn't find any reference to these facts.