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No, the matrix $$ \begin{pmatrix}0 & 1 & 0\\ 0 & 0 & p\\ 0 & 0 & 0 \end{pmatrix}\in\mathrm{M}_{3}(\mathbb{Z}_{p}) $$

is not similar to its transpose. This has been known for some time. McDonald, Linear Algebra over Commutative Rings (1984) atttributes this to M. Hochster in Exercise V.D.17, p. 424. More generally, it is noted there that this counter-example works for any commutative ring $R$ that is not von Neumann regular when instead of $p$ one uses an element $x\in R$ such that $x\not\in (x^2)$.

The "reason" why there is no counter-example for $2\times 2$ matrices over $\mathbb{Z}_p$ is that any such matrix can be written as $aI+p^iX$, where a is a scalar and $X$ is a regular matrix, that is, $X$ is similar to a companion matrix. The point is that it is known that companion matrices over commutative rings are similar to their transpose (I think this is due to Gustafson).

No, the matrix $$ \begin{pmatrix}0 & 1 & 0\\ 0 & 0 & p\\ 0 & 0 & 0 \end{pmatrix}\in\mathrm{M}_{3}(\mathbb{Z}_{p}) $$

is not similar to its transpose. This has been known for some time. McDonald, Linear Algebra over Commutative Rings (1984) atttributes this to M. Hochster in Exercise V.D.17, p. 424. More generally, it is noted there that this counter-example works for any commutative ring $R$ that is not von Neumann regular when instead of $p$ one uses an element $x\in R$ such that $x\not\in (x^2)$.

The "reason" why there is no counter-example for $2\times 2$ matrices over $\mathbb{Z}_p$ is that any such matrix can be written as $aI+p^iX$, where $a$ is a scalar and $X$ is a regular matrix, that is, $X$ is similar to a companion matrix. The point is that it is known that companion matrices over commutative rings are similar to their transpose (I think this is due to Gustafson).

No, the matrix $$ \begin{pmatrix}0 & 1 & 0\\ 0 & 0 & p\\ 0 & 0 & 0 \end{pmatrix}\in\mathrm{M}_{3}(\mathbb{Z}_{p}) $$

is not similar to its transpose. This has been known for some time. McDonald, Linear Algebra over Commutative Rings (1984) atttributes this to M. Hochster in Exercise V.D.17, p. 424. More generally, it is noted there that this counter-example works for any commutative ring $R$ that is not von Neumann regular when $p$ is replaced by an element $x\in R$ such that $x\not\in (x^2)$.

The "reason" why there is no counter-example for $2\times 2$ matrices over $\mathbb{Z}_p$ is that any such matrix can be written as $aI+p^iX$, where a is a scalar and $X$ is a regular matrix, that is, $X$ is similar to a companion matrix. The point is that it is known that companion matrices over commutative rings are similar to their transpose (I think this is due to Gustafson).

No, the matrix $$ \begin{pmatrix}0 & 1 & 0\\ 0 & 0 & p\\ 0 & 0 & 0 \end{pmatrix}\in\mathrm{M}_{3}(\mathbb{Z}_{p}) $$

is not similar to its transpose. This has been known for some time. McDonald, Linear Algebra over Commutative Rings (1984) atttributes this to M. Hochster in Exercise V.D.17, p. 424. More generally, it is noted there that this counter-example works for any commutative ring $R$ that is not von Neumann regular when instead of $p$ one uses an element $x\in R$ such that $x\not\in (x^2)$.

The "reason" why there is no counter-example for $2\times 2$ matrices over $\mathbb{Z}_p$ is that any such matrix can be written as $aI+p^iX$, where a is a scalar and $X$ is a regular matrix, that is, $X$ is similar to a companion matrix. The point is that it is known that companion matrices over commutative rings are similar to their transpose (I think this is due to Gustafson).

No, the matrix $$ \begin{pmatrix}0 & 1 & 0\\ 0 & 0 & p\\ 0 & 0 & 0 \end{pmatrix}\in\mathrm{M}_{3}(\mathbb{Z}_{p}) $$

is not similar to its transpose. This has been known for some time. McDonald, Linear Algebra over Commutative Rings (1984) atttributes this to M. Hochster in Exercise V.D.17, p. 424. More generally, it is noted there that this counter-example works over any commutative ring that is not von Neumann regular.

The "reason" why there is no counter-example for $2\times 2$ matrices over $\mathbb{Z}_p$ is that any such matrix can be written as $aI+p^iX$, where a is a scalar and $X$ is a regular matrix, that is, $X$ is similar to a companion matrix. The point is that it is known that companion matrices over commutative rings are similar to their transpose (I think this is due to Gustafson).

No, the matrix $$ \begin{pmatrix}0 & 1 & 0\\ 0 & 0 & p\\ 0 & 0 & 0 \end{pmatrix}\in\mathrm{M}_{3}(\mathbb{Z}_{p}) $$

is not similar to its transpose. This has been known for some time. McDonald, Linear Algebra over Commutative Rings (1984) atttributes this to M. Hochster in Exercise V.D.17, p. 424. More generally, it is noted there that this counter-example works for any commutative ring $R$ that is not von Neumann regular when $p$ is replaced by an element $x\in R$ such that $x\not\in (x^2)$.

The "reason" why there is no counter-example for $2\times 2$ matrices over $\mathbb{Z}_p$ is that any such matrix can be written as $aI+p^iX$, where a is a scalar and $X$ is a regular matrix, that is, $X$ is similar to a companion matrix. The point is that it is known that companion matrices over commutative rings are similar to their transpose (I think this is due to Gustafson).